A LGEBRA
IN
B USINESS AND E CONOMICS
G. S. Monk
Revised Autumn 2009
Published by Professional Copy
4200 University Way NE,
Seattle, WA 98105
(206-634-2689)
Not to be reproduced without authors’ permission.
ii
About Math 111
Math 111 will probably be very different from most Math courses you’ve previously experienced. It doesn’t follow the example/problem/example/problem model where you simply work
problems exactly like given examples, only with different numbers. There are very few drill problems. You’ll often be asked questions which need to be answered with words instead of numbers
or formulas. You’ll be challenged to think about ideas rather than plugging numbers into formulas.
This makes some people uncomfortable at first, but most people find that they get used to it with
practice.
The course introduces several techniques (especially in the realm of working with graphs using
rulers) which might not be familiar to you. It also uses some language (terms from business and
economics as well as some idiosyncratic vocabulary) which might be new to you.
We can’t overemphasize the importance of reading and thinking about all of the worksheets
and all of the questions on each worksheet. It often isn’t reasonable to skip over some parts - the
worksheets really do build on each other.
How to Use This Text
This isn’t a traditional textbook. Rather than a lot of exposition, the main ideas of the course
are communicated through worksheets. Each worksheet begins with one or more Key Questions.
The Key Questions are followed by a list of exercises.
You should begin each worksheet by reading and thinking about the Key Questions. You won’t
always be able to answer the Key Questions right away. That’s what the exercises are for. They will
lead you to an understanding of the issues involved and, we hope, to answers to the Key Questions.
Each exercise is numbered, like 8 . You should work all of the exercises carefully. Sometimes, you might not see right away what it is that they’re getting at but go ahead and play along
anyway. Write up your solutions to ALL of the exercises. Don’t just jot down computations,
really write up solutions. We’ve included an appendix that contains graphs and tables that you can
add to and fill in and then tear out to include in your homework assignments.
Interspersed with the exercises, you’ll find bits of exposition. They will be formatted like this paragraph.
They don’t require answers, but you’ll certainly want to read them.
Most Worksheets in this text will end with a few exercises marked with an → . These exercises
are intended to test your understanding of the material contained in the worksheet and, in some
cases, extend the work you’ve already done. Some of these exercises will be quite challenging,
but be persistent and seek help when you need it. The arrow exercises given at the end of the
worksheets should be a good indication of the type and difficulty of exam questions.
Math Study Center (MSC)
Because of the challenging nature of this course, the Mathematics Department offers a Study
Center for the students in Math 111 and Math 112. The Math Study Center provides a supportive
atmosphere for you to work on your math either individually or in groups. Details regarding the
time and location of the Math Study Center will be announced in class and on your instructor’s
website.
Comments, Suggestions, Typos, . . .
This version of the text was revised from a version that contained substantial contributions
from Ken Plochinski and Patrick Averbeck. Special thanks to Kelly Jabbusch for the Glossary and
to Matthew Conroy for the Index. This text is a work in progress. We have made very few changes
since the last printing. However, you may still find mistakes here and there. If you find any errors,
please report them via email to
[email protected]
Also, feel free to send in your comments and suggestions. We will keep an updated list of corrections posted on the web at
http://www.math.washington.edu/∼m111/m111typos.html
Contents
Prologue: Topics for Review
Straight Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
5
8
Information from Graphs and Tables
Worksheet #1.
Worksheet #2.
Worksheet #3.
Worksheet #4.
Introduction to Rates of Change
Rates of Flow at a Reservoir . .
Print Shop . . . . . . . . . . . .
Increments and Speeds . . . . .
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15
23
28
32
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39
43
50
56
61
Functional Notation and Graphs
Worksheet #5.
Worksheet #6.
Worksheet #7.
Worksheet #8.
Worksheet #9.
The Lagging Car . . . . . . . .
A Reservoir in Three Languages
Increments and Reference Lines
Analysis of Costs I . . . . . . .
Analysis of Costs II . . . . . . .
Linear Analysis
Worksheet #10. Breaking Even . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Worksheet #11. A Car with a Formula . . . . . . . . . . . . . . . . . . . . . . . . 73
Worksheet #12. The Changing Price List at the Print Shop . . . . . . . . . . . . . 80
Quadratic Analysis
Worksheet #13.
Worksheet #14.
Worksheet #15.
Worksheet #16.
Two Vats of Water . . . . . . . . . .
The Area-of-the-Rectangle Function
The Analysis of Costs III . . . . . .
Distances from Speed Formulas . . .
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89
98
105
110
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119
127
130
135
140
147
150
Exponential and Logarithmic Functions
Worksheet #17.
Worksheet #18.
Worksheet #19.
Worksheet #20.
Worksheet #21.
Worksheet #22.
Worksheet #23.
Growth of Sequences . . . .
The Inheritance . . . . . . .
Going Backwards in Time . .
Partial Compounding Periods
Interest . . . . . . . . . . . .
Annual Percentage Yield . .
Solving for Time . . . . . . .
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Index
Appendices
155
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
Graphs and Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Prologue: Topics for Review
1
S TRAIGHT L INES
3
Straight Lines
y
The slope of a line measures its inclination or steepness. Think of yourself as moving
along the line shown at right, from the point
P to the point Q. As you do so, you will go in
the x-direction from the value a to the value
c. Thus you will go across by the distance
(c − a). This is called the “run.” At the same
time you will go in the y-direction from the
value b to the value d. Thus you will go up by
the distance (d − b). This is called the “rise.”
.
.
......
Q : (c, d) ............
..•......
d ...
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..
.
..
........ ↑|
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..
.
rise = (d − b)
..
........
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.
.
..
.
P : (a, b) ......
.. −−− run −−−−→↓|
.
....←−
.
.
•
b ...
.
.
.
.
.
.
..
.....
= (c − a)
.. ............
.. ........
.
.............
..
..
............................................................................................................................................................................
..
a
c
x
The slope of the line you are traveling along is then defined to be the quotient:
Slope =
rise
(d − b)
=
.
run
(c − a)
One of the most important facts about slopes of straight lines is the following: if you are measuring
the slope of a given straight line, it does not matter what pair of points on the line you use for your
calculations.
The best way to see this is to calculate the slope of the line shown in the diagram to the left below. First
calculate the slope going from A to D, and then calculate the slope going from B to C. Regardless of which
two points you use to compute it, the slope should always be 2.
.
.....
.....
D..........
..
..
...•
C ......... (5, 13)
..
..
......•(4, 11)
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.
..
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.
.
..
.....
..
B .........
..
....•(2, 7)
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..
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.
.
.
.. .......
A............
..•... (0, 3)
..
.....................................................................................................................................
.
.
.....
.....
D..........
..
..
...•
C .........
..
..
......•
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..
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.
..
.....
..
B .........
..
•
....•
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.
..
.
P
.
.
.
.. .......
.
.
.
A.........
•
..•...
Q
..
.
....................................................................................................................................
.
The reason it does not matter what points you choose to measure slopes from is shown in the diagram to
the right above. The △AQD is similar to the △BP C. Therefore, from geometry we have QD : AQ =
P C : BP . Another way of saying this is that QD/AQ = P C/BP . This says that the (rise/run) is the same
in going from A to D as it is in going from B to C.
4
P ROLOGUE : T OPICS FOR R EVIEW
If you think of yourself as traveling from left to right along a straight line, then the line will have positive
slope if it goes up and negative slope if it goes down.
Positive slope..........
.
.......
.
.
.
.
.
....
.......
.
.
.
.
.
.......
.......
1
.......
.......
.......
.......
.......
.......
.......
....
Negative slope ......
Calculate the slopes of the straight lines shown below.
A
.z
...... ....
24 ... ....... .•..........
.. .. ..
.. .... ...
12 ... .... .....
....
.. ....
.. ....
...............................................................................................
... ..... a
..
.. 5 10 15 ...20
.... 25.......
−12 ...
... ....
..
......... ....... ....... ....... ....... ....... ....... ....... ..................
•...
−24 ..
.
..
2
B
y
.
............. ....... ....... ....... ....... ....... ....... ....... .....•........
2 ..
..... ........
.
..
.
.
.
.
..
..... ........
.
..
.
.
.
.
..
1 ......... ....... .......... .......•.........
...
..
.. .. .....
...
.
...
..
..
...
..
..
...
..............................................................................................
..
1 2 3 4 5 x
C
.p
......
16 ..
(10, 12)......
.
12 ...
..........•
.. ........................
8 .. ....•
(2, 8)
..
4 ..
.
............................................................................................
q
2 4 6 8 10
(Note: One of the appendices at the back of the book contains copies of graphs and tables
that you can tear out and include in your homework.)
(a) On the axes “A” above draw a line through (5, 24) with slope −12/5.
(b) On the axes “B” draw the line that goes through (0, 1) and (4, 3/2). Calculate its slope.
(c) On the axes “C” draw a line that goes through the origin and has slope 1.
3
Is the point (10, 12) on the line shown on “A”? HINT: If (10, 12) is on the line, then the slope
from (5, 24) to (10, 12) should be the same as the slope from (5, 24) to (25, −24).
4
Is the point (3, 43 ) on the line shown in “B”?
L INEAR F UNCTIONS
5
Linear Functions
If the formula for a function looks like
f (x) = mx + b,
then the graph of the function is a straight
line, and the function is called a linear function. The number m in the formula is the
slope of its straight-line graph and the number b is its y-intercept, the y-coordinate of
the point at which the line crosses the vertical axis.
Conversely, if a function has a straight-line
graph, then it is given by a formula of the form
f (x) = mx + b.
The two graphs at right are of the linear
functions:
f (x) = 2x + 3,
g(x) = −x + 9.
y
.
......
..
f (x) = 2x + 3........
12 ...
....
.
.
.
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.
..
......
..
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...
.
....
9 ................
.............. .............
..
..
.................
..
...... .........................
.
.
6 ...
.
.
.
.
..............
..
....
........
.. .........
.......
g(x) = −x + 9
3 ...
..
..
.
.........................................................................................................................................................
x
..
1
2
3
4
5
Given a linear function, you can sketch its graph using either of the methods illustrated as follows:
i) Since you know that the graph of f (x) = 2x + 3 is a straight line, all you need is two points, and
then you can draw the line. You can get these two points by plugging in any pair of values, e.g.,
f (2) = 2 · 2 + 3 = 7; f (5) = 2 · 5 + 3 = 13. Then plot (2, 7) and (5, 13) on the axes and draw the
straight line connecting them.
ii) You can draw the graph of the straight line using the slope and the y-intercept. The line h(x) = 3x−2
has slope 3 and y-intercept −2. This means that the line goes through the y-axis at the point (0, −2)
and, if you move to the right 1 unit and up 3 units, you will hit another point on the line. So, plot the
y-intercept and the point that is 1 unit to the right and 3 units up and draw the straight line that goes
through these two points.
5
Draw the graphs of the following linear functions.
a) f (t) = 3t + 4,
b) g(x) = 12 − 6x,
c) h(x) = 5,
d) p(t) = 21 t − 6.
6
P ROLOGUE : T OPICS FOR R EVIEW
Now let’s go the other direction: start with
the graph of a line and find its formula. Suppose you are given the graph at right and you
want the formula for the linear function that
it defines.
y
.....
12 ........
.. ....
.. ....
8 ... ....... ....... .....•...........
... ..
..
..
.....
..
...
..
..
.
4 ......... ....... ............ ....... ....... .....•............
...
..
.....
...
..
..
..
.
...
....
..
.
...........................................................................................
.
x
2
4
6
If you can accurately find the coordinates of two points on the line, then compute its slope. For the
line above, the slope is −2, as calculated from the two points (2, 8) and (4, 4). So, the formula looks like
f (x) = −2x + b, and you want to determine b. From the fact that (2, 8) is on the graph, you know that
8 = f (2) = (−2)(2) + b.
This becomes 8 = −4 + b, and we get b = 12. (In this case, it is clear from the graph that the y-intercept
is 12. But the method illustrated in this example will work when the y-intercept cannot be read accurately
from the graph.)
6
Write out the formulas for the linear functions
whose graphs are shown to the right.
y
A..
...8. ..... ...C.....
..
.
.
.... ...
.
.
.
.
.
.
...... ....
.
........
6 ........
.
.
.
.. ....
.
.. ........................B .
.
4 .. ....
.
.
. ... ................. ......
...
2 ...............................
........................................................................................................................................
.
.. ....
..... 1 2 ...3... 4 5 6 x
.
−2... ..
....
. ..
.
.... .
.
−4 ..
... D
In Math 111 and 112, you will occasionally need to solve a system of two equations in two unknowns, like
the following.
3x + 2y = 6
2x − 4y = 5.
Solving this pair of equations amounts to finding the point of intersection of the two lines represented by
these equations. Most people will use one of the following methods for solving these equations for x and y.
i) Choose one variable, solve both equations for it, and then set the results equal to one another. For
example, you can solve both of the above equations for y, and get y = (6 − 3x)/2 and y = (5 −
2x)/(−4). Then these two expressions that are equal to y must be equal to one another. Thus
6 − 3x
5 − 2x
=
,
2
−4
L INEAR F UNCTIONS
7
which becomes, if you multiply both sides by 4,
2(6 − 3x)
12 − 6x
17
x
=
−
=
=
(−1)(5 − 2x)
−5 + 2x
8x
17/8.
Then you plug this value x = 17/8 into one of the original equations to get the corresponding value
of y, e.g.,
2(17/8) − 4y = 5,
(17/4) − 5 = 4y,
y = (1/4)(−3/4) = −3/16.
Note: You can check your work by plugging your solutions into the other original equation and verifying that they work.
.....
3(17/8) + 2(−3/16) = (51/8) − (3/8) = (48/8) = 6
ii) Solve one equation for one variable and plug the result into the other equation.
Solving the first equation for y, we get y = (6 − 3x)/2. Then plug this into the second equation and
we get a single equation in x:
2x − 4(
6 − 3x
) = 5,
2
2x − 12 + 6x = 5,
8x = 17,
x = 17/8.
Again, you plug back into some equation to get y. Here you can use the equation
y = (6 − 3x)/2,
and you get
y = [6 − (3)(17)/8]/2 = −3/16.
iii) Work on both equations at the same time and eliminate a variable. The idea here is that, if you multiply
one equation by one number (cleverly chosen) and the other equation by another number (also cleverly
chosen), you will get two new equations in which a variable has the same coefficient in both equations.
This allows you to get rid of that variable. For instance, if you multiply the first equation (above) by
2 and the second equation by 3, then both of your new equations will have a 6 as the coefficient of x.
Likewise, if you multiply the upper one by 2, then the coefficient of y in the upper one will be 4, and
the coefficient of y in the lower one will be −4, which is close enough. Then watch
6x
6x
Subtract
+
-
4y
12y
16y
y
=
=
=
=
12
15
-3
-3/16
Add
6x
2x
8x
x
+
-
4y
4y
+
=
=
=
12
5
17
17/8
You don’t need to do both of these procedures. You choose one, solve for the resulting variable, and
then go back to either of the original equations to solve for the other.
7
Use any method you like to solve the following pairs of linear equations.
a)
s + 3t = 5
−2s + 3t = 6
b)
2p − q = 6
3p − q = 9
c)
x + 2 = 4y
2x − 3 = 2y
8
P ROLOGUE : T OPICS FOR R EVIEW
Quadratics
In this section you will find the basic information about functions given by formulas that
look like
f (x) = ax2 + bx + c.
Such a function is called a quadratic function.
The graph of a quadratic function is shaped
like one of the two shown to the right. This
shape graph is called a parabola.
Whether or not the graph faces up or down
depends on the particular number that precedes the x2 —this is called the coefficient of
x2 —it is the value of a in the above form.
For the upward facing parabola to the
right, the coefficient of x2 is 1. For the downward facing parabola to the right, the coefficient of x2 is −2.
y
.....
16 ...
g(x) = x2 − 6x + 8
..
..
.
14 ..
.
.
..
..................
.
.
12 ...
.
.
.
..
....
..
..
.
.
10 ...
.
.
.
... ..
.
.. ..
... ...
8 ......
.
.. ..
.
... ...
.
.
6 .. ...
...
.
.. .. ..
.
.
.
.
.
4 . ... .
. .....
..
.
.
.
.
.
.
2 ...
.. ....
.........
.
.
.
.
.
.
.
................................................................................................................................................................................
..............
x
.. ..
..
−2 ... ...1 2 3 4 5 6... 7
. .
...
−4 ... ...
...
.. ..
...
.
.
−6 . ..
.. .
...
−8 ... ...
...
..
f (x) = −2x2 + 14x − 12 ...
−10 ......
...
..
−12 .....
The general rule is:
A parabola is shaped like a mountain, or opens down, if a is negative.
A parabola is shaped like a valley, or opens up, if a is positive.
In this course, you will often be expected to apply the following formulas:
The Quadratic Formula Given f (x) = ax2 + bx + c, the values of x at which f (x) = 0 (i.e., the roots of
the equation ax2 + bx + c = 0) are given by the formula
√
−b ± b2 − 4ac
.
x=
2a
These values x are sometimes called the roots (or zeros) of the function. They are the values of x at which
the graph, the parabola, crosses the x-axis.
The Vertex Formula Given f (x) = ax2 + bx + c, the vertex is the high point, in case the parabola is
downward facing, and the vertex is the low point, in case the parabola is upward facing. The value of x at
which the graph reaches its vertex is given by the formula
x=−
b
2a
Q UADRATICS
9
y
.....
..
...
..
..
.
..
.
.
...
..
..
..
.
.
.
...
..
.
.
.
..
.
.
...
..
.... ....
..
..
..............
..
...
..
.
..
.
....
.
.......................................................................................................................
x
..
b
..
−
.
2a
y .
.
....
...... ....... ...... ........
.. .. .... ...
...
..
.
.......... ........ .....
.
...
..
...
.
...
..... .....
.
....
...
...
... ...
.
...
..
..
..
..... .....
...
.
.............................................................................................................................
. .
...
x
..... ..... − b
...
..
.. .. 2a
The actual height of the vertex can be found by plugging the x-value given by the formula above back into
the formula that gives the function. The y–coordinate of the vertex is
y=f
−b
2a
−b
=a
2a
2
−b
+b
2a
+c=
b2
b2
b2 − 4ac
b2
−
+c=− +c=−
4a 2a
4a
4a
The following examples demonstrate how these formulas are used.
Example 1. The graph of the function f (x) = −2x2 + 14x − 12 was given at the beginning of this section
on Quadratics. The roots of this function are:
√
p
−14 ± 142 − 4(−2)(−12)
−14 ± 100
−14 ± 10
x=
=
=
= 6, 1.
2(−2)
−4
−4
The graph crosses the x-axis at 6 and 1, as you can see.
The vertex of this parabola is at x = (−14)/(−4) = 3.5. You can find the height of the vertex by plugging
x = 3.5 into the formula for f (x) and getting
f (3.5) = −2(3.5)2 + 14(3.5) − 12 = 12.5.
Example 2. The graph of the function g(x) = x2 − 6x + 8 was also given at the beginning of this section.
Its roots are:
√
p
6± 4
−(−6) ± (−6)2 − 4(1)(8)
=
= 2, 4.
x=
2
2
The vertex of this parabola is at x = 3 and its height there is −1. You should check these facts on the graph.
Example 3. Next consider the function h(x) = x2 − 6x + 21. Since the coefficient of x2 is 1, the parabola
is upward facing. The roots of this function are at
p
√
6 ± −48
−(−6) ± (−6)2 − 4(1)(21)
=
.
x=
2
2
But now we have a problem. There is no (real) square root of −48. You can’t take square roots of negative
numbers. This simply means that the function has no roots and that the parabola never crosses the x-axis.
10
P ROLOGUE : T OPICS FOR R EVIEW
To get some idea of what the parabola looks like, we locate its vertex. This is at x = 3 and y = 12. Thus
the parabola looks like
y
..... .
.. .....
....
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.......
.
..
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..
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.....
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12 .......... ....... ....... ....... ....... ....... ...............................................
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.
.........................................................................................................................................................................
x
3
Example 4. Finally consider the quadratic function p(x) = −x2 + 8x − 16. The coefficient of x2 is −1, so
that the parabola is downward facing. In view of example 3, we try to find roots for the function (we are no
longer certain that there are any). Then we get
x =
−8 ±
q
64 − 4(−1)(−16)
√ −2
−8 ± 0
=4
=
−2
That was simple. But we only got one root, not our customary two. What does that mean? It only means that the
parabola looks like the one to the right. It touches, but
doesn’t actually cross the x-axis.
8
9
y
.....
..
..
.........................................................................................................................................................................
.
x
..
..... 4 .........
..
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.
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..
..
By using the quadratic formula, solve each of the following quadratic equations.
a) 3x2 + 9x − 30 = 0,
b) − 21 t2 + 72 t − 6 = 0
c) q 2 + 25 q − 6 = 0,
d) 2x2 − 6x + 9 = 0
Solve each of the following quadratic equations by first putting it in the same form as those
in the previous exercise, and then using the quadratic formula.
a) 2r2 − 6r − 6 = −2
b) − 12 t2 + 27 t − 6 = 12
c) 3x2 + 6x − 20 = 10 − 3x
d) x2 − 4x + 8 = 2x + 6
e) −x2 + 6x − 2 = x2 + 2x + 1
f) t2 − 2t + 1 = 2t2 + 4t − 5
Q UADRATICS
11
10 Using information gathered in exercise #8, and then the vertex formula, graph the following
quadratic functions.
a) f (x) = 3x2 + 9x − 30
b) g(t) = − 21 t2 + 72 t − 6
c) h(q) = q 2 + 25 q − 6
d) p(x) = 2x2 − 6x + 9.
In exercise #9 above, you were given equations which did not have 0 on one side, and you were to first
transfer the terms over, so that you got an equation with 0 on one side. However, there is another way of
looking at this problem.
With the equations taken as written, you can look at each
side of an equation as giving a function, and the equation
asks for when the two functions are equal, which is the
same as asking when their graphs cross, e.g., in (a) the
left-hand side defines the function f (r) = 2r2 −6r−6, and
the right hand side defines the function g(r) = −2, and
you are asking for when the graphs cross. By setting the
two expressions equal to one another, and solving, as
√ you
6± 68
did in #9(a), we find that the graphs cross at r = 4 .
Similarly, we can use the equation in #9(c) to give us two
functions
A(x) = 3x2 + 6x − 20
B(x) = 10 − 3x.
These are shown to the right. Your solutions to the equation #9(c) tell us that the two graphs cross at x = 2 and
x = −5.
y
.....
.
...
..
..
.
..
...
.
..
... 10 ....
.
.
... ...
....
... ..
.
... ...
..
.
.
.
......................................................................................................................................
.•.. ...
r
............................
.. •.. 4
2
.....
.
.......
..
.
.. ....
.
. ..
...
−10 ... ................
.
.y
...
......
..
......•.....
..
..
... ...........
.
.
.
.
... .............20 ...
....
............
...
.
............ ..
...
.. ......•....
.
..........................................................................................................
x
.
.. ..
.. .. 2
−5 ....
.. ..
...
.. ..
....
.−20
.....................
11 Each of the equations in exercise #9 tells us about the intersection of two graphs. Sketch
each of the two graphs (the one defined by the left-hand side of the equation and the one
defined by the right-hand side of the equation), and then use your solutions to exercise #9 to
indicate where these two graphs intersect.
12
P ROLOGUE : T OPICS FOR R EVIEW
Information from Graphs and Tables
13
Worksheet #1
Introduction to Rates of Change
Key Question
You are traveling along the freeway on a road trip. After passing some construction work, you
begin to speed up. Does this necessarily mean that the average speed for your trip is also going
up? Or is it possible for your speed to go up while your average trip speed goes down?
Speed is an example of a rate of change. You may be familiar with the formula
speed =
distance
.
time
Speed is the change in distance per unit time. That is, speed is the rate at which distance changes. We
measure speed in units like miles per hour or feet per second or inches per year, and so forth.
There are two types of rate of change you will encounter in this course: overall and incremental. An
overall rate of change is a rate of change measured from the starting time to a later time. An incremental
rate of change is a rate of change measured between any two times.
In the context of speed, for example, if you drive 60 miles in the first 60 minutes of your trip, then your
average trip speed at time t = 60 minutes is 1 mile per minute. Since this speed is computed using the
distance covered and time elapsed since the beginning of your trip, average trip speed is an example of an
overall rate of change. On the other hand, if you time a mile in the middle of your trip and find that it takes
1.6 minutes, then your average speed over that period of time is
1 mile
mile
1
=
= 0.625 miles per minute.
1.6 minutes
1.6 minutes
Since this average speed is not measured since the beginning of your trip, this is an incremental rate of
change. (The actual speed of the car at any instant is an example of an instantaneous rate of change,
which we will cover in Math 112.)
1
Use your intuition to answer the Key Question.
In order to deal with the Key Question more specifically, we have a graphical representation of the distance vs. time for the first hour of the trip.
15
16
W ORKSHEET #1
48
44
40
Distance (miles)
36
32
28
24
20
16
12
8
4
2
Introduction to Rates of Change
.
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...................................................................................................................................................................... Time
10
20
30
40
50
60 (minutes)
To confirm that you are reading the graph appropriately, determine the following.
(a) How far does the car travel in the first 30 minutes?
(b) How far does the car travel from time 40 minutes to time 60 minutes?
3
In the first five minutes, the car travels 8.5 miles. To calculate the average trip speed at
t = 5 minutes, we divide the 8.5 miles by 5 minutes and get 1.70 miles per minute (mpm).
Use this process to fill in the following table with the appropriate average trip speeds, i.e.,
under the time t minutes, put the average speed from the start of the trip to t minutes.
t
AT S
0
N/A
5
1.70
10
1.46
15
1.25
20
25
30
0.80
35
40
0.67
45
50
55
60
t = time in minutes, AT S = average trip speed in mpm
4
Using the data in the table above, determine which of the following statements is true:
(a) The average trip speed always increases.
(b) The average trip speed always decreases.
(c) The average trip speed increases then decreases.
(d) The average trip speed decreases then increases.
W ORKSHEET #1
Introduction to Rates of Change
17
Recall that the Key Question asks you to compare the average trip speed to the actual speed. Computing
actual speeds from the graph we’ve been given is a difficult task that we’ll put off until Math 112. For
now, we’ll compute average speed over a small time interval and use this average speed to approximate
the car’s actual speed. For example, during the 5-minute interval from t = 35 to t = 40, the car travels
approximately 1.6 miles. We say that, on this interval, the change in distance is 1.6 miles and the change in
time is 5 minutes. In shorthand, we write ∆D = 1.6 miles and ∆t = 5 minutes. The car’s average speed
over this interval is:
1.6 miles
∆D
=
= 0.32 mpm.
∆t
5 minutes
This triangle ∆ is the Greek letter delta. It is often read “the change in.” For example, when you see the
mathematical sentence ∆D = 1.6 miles, you should say “the change in distance is 1.6 miles.”
5
Which of the following would you choose as the better estimate of the actual speed of the
car at t = 40 minutes? Explain your choice.
(a) the average trip speed at time 40 minutes (0.67 mpm)
(b) the average speed from t = 35 to t = 40 minutes (0.32 mpm)
6
Using the above method for calculating average speed over an interval, complete the following table. Notice that, on each of these intervals, ∆t = 5 minutes.
TI
0–5 5–10 10–15 15–20 20–25 25–30 30–35 35–40 40–45 45–50
∆D
8.5
6.1
4.1
1.0
1.1
4.0
AS
1.70 1.22
0.82
0.36
0.32
T I=time interval, ∆D=change in distance over that interval, AS = average speed over interval
50–55
55–60
1.22
7
Since 5 minutes is a relatively short period of time, the average speeds that you just computed
can be used to approximate the car’s actual speed at various times. For example, we could
say that, at t = 12 minutes, the car’s actual speed is approximately 0.82 miles per minute,
the average speed over the interval from 10 to 15 minutes. Use the table from exercise 6 to
find the longest interval of time over which the car’s approximate speed is increasing.
8
Now that you have information about the car’s average trip speed and approximate speed,
go back and answer the Key Question. Incorporate into your explanation an interval of time
that supports your conclusion.
In the previous exercises, you used the definition of speed (change in distance over change in time) to
compute average trip speeds and approximate speeds. You can, however, also interpret speeds as slopes of
lines through the distance graph. While it may not be clear at first why you might want to do this, it will
eventually allow us to answer more easily questions about the behavior of the car’s speed over time (e.g.,
when is the car getting faster or slower?) and will give us more accuracy when we approximate speeds from
a graph.
The following are blow-ups of portions of the graph of distance vs. time.
18
W ORKSHEET #1
D
D
24.0
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Distance
..
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covered
.. ..
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|
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|
↓
..
t
30
←
− Time elapsed −
→
Average Trip Speed
Introduction to Rates of Change
24.0
23.0
.
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.
........
...... Distance
.
.
.
.
covered
.
.....
Time
elapsed
..............................................................................................................................
25
30
t
Average Speed
These blow-ups say:
“To get average trip speed at t = 30 from the graph, you divide the change in the height of the graph since
24
the beginning by the amount of time elapsed since the beginning. (Average trip speed at t = 30 is 30
or 0.80
miles per minute.)”
“To get average speed, you divide the change in the height of the graph by the time elapsed. (Average speed
from t = 25 to t = 30 is 15 or 0.20 miles per minute.)”
9
On the blow-up of the average trip speed illustration, draw a line connecting the origin
(0, 0) with the point (30, 24) and compute the slope of this line. On the blow-up of the
average speed illustration, draw a line connecting the two points (25, 23.0) and (30, 24.0)
and compute the slope of this line. How are the concepts of distance covered and time
elapsed related to the slope concepts rise and run?
All the work that you did in order to answer the Key Question was directly related to slopes. A secant line
is a straight line connecting two points on a graph. The average speed from t = 25 to t = 30 is the slope
of the secant line that goes through the distance graph at t = 25 and t = 30 (one of the lines you drew in
the previous exercise). The average trip speed at t = 30 is the slope of the secant line that goes through the
distance graph at t = 0 and at t = 30 (the other line you drew in the previous exercise). Any line through
the origin is called a diagonal line. If your distance graph goes through the origin, then the line whose
slope gives average trip speed is both a secant line and a diagonal line.
10 Helpful hint: To measure the slope of a straight line it doesn’t matter which two points on the
line you use. So first draw the secant line using the correct two points, but then for accuracy,
extend the line in both directions and use the two points that are most convenient to get rise
and run.
(a) On the original graph of distance vs. time, draw three diagonal lines corresponding to
the times t = 20, t = 30, and t = 40 minutes. Measure their slopes. Check your results
against the table readings in exercise #3.
(b) Draw secant lines corresponding to the following three time intervals: between 5 and
10 minutes, between 25 and 30 minutes, between 45 and 50 minutes. Measure their
slopes. Check your results against the table readings in exercise #6.
W ORKSHEET #1
Introduction to Rates of Change
19
At this point, you should be able to identify the relationship between the two speeds and the slopes of the
two lines we’ve encountered. Since the graph of distance goes through the origin, the average trip speed
(overall rate of change of distance) is represented graphically by the slope of a diagonal line. Other average
speeds (incremental rates of change of distance) are represented graphically by slopes of secant lines. We’re
ready to use the graph to answer the Key Question yet again.
Place your ruler so that it goes through the origin and through the distance graph at t = 5. (If you were
to draw a line here, its slope would be the car’s average trip speed at t = 5.) Now “roll” the ruler so that
one end stays on the origin and the other follows the curve of the distance graph. Notice that, as you move
from t = 5 to t = 15, for example, the diagonal lines get less steep. The slopes get smaller. (What does this
mean about average trip speeds?)
11 Continue to move your ruler along the graph. Determine the time at which the average trip
speed is lowest. (Confirm your answers with the table in exercise #3.)
A similar graphical method can be used to examine the behavior of average speeds over five-minute
intervals.
12 Use your ruler to visualize the secant line between t = 5 and t = 10. (You don’t need to draw
the line. Just place your ruler as if you were going to draw the line.) Then, move your ruler
to visualize the secant line between t = 10 and t = 15. Notice that the ruler gets less steep.
Next, move your ruler to visualize the secant line between t = 15 and t = 20. Once again,
the ruler gets less steep. Continue to move your ruler along the graph, showing secant lines
over 5-minute time intervals. Determine the 5-minute interval over which the approximate
speedometer reading (the average speed over that interval) is the lowest. (Confirm your
answer with the table in exercise #6.)
At this point you should understand the following:
• Average trip speed is an example of an overall rate of change. On a graph, you “see” the average trip
speed as the slope of a secant line through the distance graph at time 0 and at time t. (If the distance
graph goes through the origin, then this secant line is also a diagonal line.)
20
W ORKSHEET #1
Introduction to Rates of Change
• We can also compute the average speed over a more general interval, from time t = a to time t = b.
This is an example of an incremental rate of change. The average speed from t = a to t = b is the
slope of the secant line through the distance graph at t = a and at t = b. (Note: Average trip speed is
just a special case of the average speed.)
• The “rolling ruler” method allows you to investigate the behavior of these two rates of change.
→
13 The graph below is of the distance vs. time for a different car.
d (in. miles)
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4 .. ..
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2 ... ...
.......................................................................................................................................................................................................
5 10 15 20 25 30 35 40 45 50 55 60
t (in minutes)
a) How far did the car go from t = 20 minutes to t = 45 minutes?
b) Find the average trip speed at time 25 minutes.
c) Approximate the speedometer reading at t = 25 minutes by computing the average
speed over the five-minute interval that begins at t = 25.
d) Give a 5-minute interval over which the car travels at least 2 miles, but the average trip
speed for any time in the interval is less than 0.4 mpm.
e) Find a time at which the average speed over the next 5 minutes is equal to the average
speed over the 5-minute interval starting at t = 15.
W ORKSHEET #1
Introduction to Rates of Change
21
f) Compute the average speed from t = 42.5 to t = 47.5. Use this as an approximation
of the speedometer reading over this interval and approximate how far the car travels
from t = 45 to t = 47.
g) Find a 5-minute interval during which the change in distance is 1 mile.
h) Find the lowest value of the average trip speed for the car over the 60 minutes shown.
→
14 Two cars are in a race starting at the
same point. To the right is the graph
of distance (in miles) between the
cars at the various times in the race.
When the graph reads a positive
number, this means that Car A is
ahead. When the graph is negative,
Car B is ahead.
a) If Car B travels 40 miles in
the first hour, how far does
Car A travel in that time?
b) Which car traveled furthest in
the fourth hour (i.e. from
180 to 240 minutes)? Explain
your answer.
5.0
2.5
0
60
120
180
240
t (minutes)
-2.5
-5.0
c) Name a time when Car B is 1.75 miles ahead of A, but A is catching up.
d) Suppose B is 63 miles from the start at 180 minutes. How far from the start is Car A
at that time?
e) Tell whether the following statement is true, false, or you have no way of knowing.
“If A is standing still from t = 90 to t = 120, then B must be going backwards.”
15 The graph below is of the average trip speed of a race car for a 30 minute period.
a) Find a time when the average trip
speed of the car is 2.75 mpm.
b) How far has the car traveled during
the first 10 minutes?
c) How far does the car travel between
the two times at which its average
trip speed is 2.5 mpm?
Average Trip Speed (mpm)
→
4
3
2
1
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..
5
10
15 20
(mins)
25
30
22
W ORKSHEET #1
Introduction to Rates of Change
d) What is the average (incremental) speed of the car over the time period t = 15 to
t = 25 minutes?
e) Which of the graphs below could be the graph of distance vs. t over the interval t = 15
to t = 25 minutes? Give reasons for your answer.
..
.
.
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.. ....
(A)
...
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.
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.......
(B)
(C)
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15
15
15
25
25
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25
Worksheet #2
Rates of Flow at a Reservoir
Key Question
The graph below shows the amounts of water drawn from a city water reservoir by customers
over half-hour intervals from noon to 6 PM. Suppose water flows into the reservoir at the constant
rate of 2 thousand gallons per half hour (which is more than the amount used over most half-hour
intervals). If the reservoir were empty at noon, could the customers be supplied with their water
for the whole day? If not, what would be a reasonable amount of water to have in the reservoir to
be sure that the customers can be supplied with water?
5.0
4.5
Thousands of gallons
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
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•
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•
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•
•
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•
•
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•
•
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• • •
..
..
.....................................................................................................................................................................
.
1
2
3
4
5
6
Time (hours since noon)
1
This type of graph can be tricky to read. Each dot on the graph shows the amount of water,
in thousands of gallons, drawn from the reservoir during the half-hour interval ending at the
corresponding time. For example, the point with coordinates (1, 2.5) indicates that, between
12:30 and 1:00 (the half-hour interval ending at t = 1), 2.5 thousand gallons (or 2500 gallons) of water were drawn from the reservoir. Write out in one English sentence what the
point with coordinates (3.5, 0.63) indicates.
2
How much water is used in the first hour? How much water is used by customers in the time
interval from 1:30 to 4:00? (Do your best at approximating the height of each dot.)
23
24
W ORKSHEET #2
3
Rates of Flow at a Reservoir
In the table below, fill in the row labeled O with the total amount of water (in thousands of
gallons) that would flow out of the reservoir from noon to time t if there was enough water.
Then, recall that water is flowing into the reservoir at a rate of 2 thousand gallons per half
hour. In the row labeled I, fill in the total amount of water (in thousands of gallons) that has
flowed into the reservoir from noon to time t.
t
O
I
0.0
0.0
0.0
0.5
3.6
2.0
1.0
6.1
1.5
7.7
2.0
2.5
3.0
3.5
10.46
4.0
11.46
4.5
5.0
15.56
5.5
6.0
t = hours since noon
O = total amount of water that flows out from noon to time t
I = total amount of water that flows in from noon to time t
Thousands of gallons
4
On the axes below, use the values you computed in exercise #3 to plot O vs. t, connecting
the points with a nice smooth graph. On the same set of axes, plot I vs. t. Clearly label each
curve.
.......
24 ....
..
22 ...
.
20 ...
..
18 ...
..
16 ...
..
14 ...
.
12 ...
..
10 ...
..
8 ...
..
6 ...
.
4 ...
..
2 ....
..
............................................................................................................................................................................................................................................................................
1
2
3
4
5
6
Time (hours since noon)
5
Answer the first part of the Key Question: if the reservoir were empty at noon, could the
customers be supplied with their water for the whole day? Explain how you know.
6
You see a water shortage whenever the Out graph is above the In graph. At what time during
the shortage is the vertical distance between the graphs the greatest? What is the vertical
distance between the two graphs at that time? How much water should be in the reservoir at
noon to ensure that the customers are supplied with water? (This is the answer to the second
part of the Key Question.)
W ORKSHEET #2
Rates of Flow at a Reservoir
25
Let’s summarize what you’ve done so far.
• You were given the water usage in half-hour blocks and you used that information to compute the total
amount of water used since noon at several times.
• You were given a verbal description of how water flowed into the reservoir and you used that description to compute the total amount of water available at several times.
• You graphed the amount that has flowed out and the amount that has flowed in.
• You found the largest vertical distance between those two graphs during a shortage to find the smallest
amount of water needed in the reservoir at noon in order to supply water all day.
7
Use the technique summarized above to answer the following questions about the reservoir
described in the Key Question.
(a) What if water flows into the reservoir at the constant rate of only 1.5 thousand gallons
per half hour? Use graphs to determine the smallest amount of water needed in the
reservoir at noon so that no shortage occurs from noon to 6 p.m.
(b) What if water flows into the reservoir at the constant rate of only 1 thousand gallons
per half hour and there are 7 thousand gallons in the reservoir at noon? Use graphs
to determine if additional water is needed in the reservoir at noon so that no shortage
occurs from noon to 6 p.m. If additional water is needed, use the graphs to determine
how much.
→
8
Barometric pressure (in inches) is
monitored over several days. The
graph to the right depicts the
CHANGES in barometric pressure
over 1/4-day intervals. Each of the
data points was placed at the beginning of the time interval. For example, the point at t = 1 represents
the change in pressure (∆P ) from
t = 1 to t = 1.25. Time t = 0
corresponds to Midnight, January
9, 1980. At t = 0, the barometric
pressure measures 27 inches.
0.3
0.2
0.1
−0.1
−0.2
−0.3
∆P
....
•
..
..
•
..
•
•
..
...
..
•
..
..............................................•.............................................•......................................
.
•
..
t
..
•
1
2
3
4
.. •
•
.
•.... •
..
•
..
..
.
••
a) Name all the periods of time over which the graph of pressure is decreasing.
b) Which of the following most closely resembles the graph of pressure from t = 2.0 to
t = 2.5?
26
W ORKSHEET #2
Rates of Flow at a Reservoir
c) For each of the following pairs of times, pick the time when the pressure is higher:
(i) t = 0.5 or t = 1
(ii) t = 1.5 or t = 2
(iii) t = 3.5 or t = 4.
d) Weather is often at its worst when the pressure is falling rapidly. When, during the
4-day period pictured above would you have expected the worst weather?
e) What is the change in pressure on January 11 (from t = 2 to t = 3)?
f) What was the change in barometric pressure on January 10?
9
Starting on his 12th birthday, Juan’s
parents measure his height every
3 months. However, rather than
record his height, they record his
incremental rate of change over the
preceding 3 months. That is, they
subtract to find out how much he
has grown in the 3 months, and then
they divide by 3 to get a monthly
growth rate. The graph to the right
gives their record of Juan’s monthly
growth rate at 3-month intervals.
a) How many inches did Juan
grow in the three months preceding his 13th birthday?
.4
GROWTH RATE (IN/MO)
→
.3
.2
.1
t
12
24
(MONTHS)
36
b) Assume that Juan was 64 inches tall on his 14th birthday. How tall was he 6 months
later?
c) Did Juan grow more in the period from his 12th to 13th birthday, or the period from his
14th to his 15th birthday?
d) There are two 3-month intervals when Juan’s growth rate is approximately 0.275 inches
per month. Which of the following statements is true of these two intervals?
(i)
(ii)
(iii)
(iv)
He is growing faster over the one interval than the other.
He is shorter over the one interval than the other.
One interval is before he reaches his peak height and the other is after.
None.
W ORKSHEET #2
Rates of Flow at a Reservoir
27
e) To calculate Juan’s lifetime growth rate at any particular time, his parents divide his
height at that time by the number of months he has been alive. Again, under the assumption that Juan is 64 inches tall on his 14th birthday, tell what Juan’s lifetime growth
rate (in/mo) was 6 months after his 13th birthday.
10 A town is using water from a reservoir that is being refilled with a system of aquaducts. The
graph below shows the total water drawn from the reservoir over the course of a day, starting
at midnight.
5000
4500
4000
total water usage (gallons)
→
3500
3000
2500
2000
1500
1000
500
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
hours after midnight
(a) Name a time when the overall average rate of water usage was 150 gallons per hour.
(b) Suppose the reservoir was empty at midnight and was being filled by the aquaduct at
a constant rate. How small could that rate be and provide enough water for the town
during this 24 hour period?
(c) Suppose that the reservoir had 3000 gallons at midnight and was being filled by the
aquaduct at a constant rate. How small could that rate be and provide enough water for
the town during this 24 hour period?
(d) Suppose the aquaduct was filling the reservoir at a constant rate of 100 gallons per hour.
How much water would there have to be in the reservoir at midnight for the town to get
all the water it needed?
Worksheet #3
Print Shop
Key Question
You own a print shop, and, in order to attract larger orders, you have devised a sliding price scale.
You charge $15 for an order of one ream (1 ream contains 500 pages), $14 per ream for an order of
two reams, $13 per ream for an order of 3 reams, and so forth. Your paper supplier charges you $4
per ream. Your partner says that you cannot run a business by charging less for more, and thinks
that you will be losing money every time a customer increases his order size. You say that as long
as you get more than $4 a ream, you’re still ahead. Who is right?
1
What is the total charge to a customer who buys 1 ream of paper? 2 reams? 3 reams?
2
Your answers to exercise #1 are values of total revenue, the total amount of money you
take in on an order. If a customer purchases q reams at a price of p dollars per ream, find
a formula for total revenue (TR) in terms of p and q. Use your formula to complete the
T R-row in the table below.
q
p
TR
TC
P
3
0
N/A
0
0
0
1
15
15
4
2
14
28
8
3
13
39
4
12
48
5
11
6
10
7
9
20
24
28
35
8
8
64
32
32
9
7
63
36
27
10
6
60
40
11
5
55
44
12
4
13
3
14
2
48
0
15
1
60
On the axes to the left below plot the graph of p vs. q (with p on the vertical axis). On the
axes to the right below plot the graph of TR vs. q (with TR on the vertical axis).
.....
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.................................................................................................................................................
.
......
..
..
..
..
..
..
..
..
..
..
..
..
..................................................................................................
.
28
W ORKSHEET #3
4
Print Shop
29
What size order will bring in the most total revenue? What is the largest possible total
revenue you can earn from one order? Is your partner correct: do you lose money every time
a customer increases his order size?
Now, we’ve determined the quantity that produces the most revenue. However, this may not be the quantity
that you want to produce. Your goal is to produce the maximum profit. The quantity that maximizes total
revenue does not necessarily maximize profit. To determine profit, we need to consider the costs involved in
running the print shop.
5
For the sake of simplicity, we’ll assume that the only cost to you in running your print shop
is the cost of paper. Total cost is the total amount of money you spend to fill an order.
Recall that paper costs you $4 per ream. Find a formula for the total cost (T C) of an order
of q reams of paper. Use your formula to complete the T C-row in the table in exercise #2.
6
Sketch the graph of T C vs. q on the same set of axes as your T R graph above. Clearly label
the two graphs so you can tell them apart.
7
Obviously, you’d like T R to be larger than T C. Use the graphs of T R and T C to decide the
largest order you’re willing to take using this price scale.
To deal further with the Key Question, we need to define profit. If you fill an order of 9 reams, then you’ll
take in $63 (your T R) and spend $36 (your T C). Your profit on this size order is $63 − 36 = $27. In
general, profit is given by the formula
P = T R − T C.
8
Now go back to the table in exercise #2 and complete the P -row. Sketch the graph of proft
vs. quantity on the axes below (with P on the vertical axis).
.
......
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..................................................................................................................................................
9
What size order must you sell in order to make the maximum profit? What is the maximum
possible profit you can earn from one order?
30
W ORKSHEET #3
Print Shop
As you can see from the table, the quantity that produces the maximum profit is 6 reams. Obtaining the
quantity from a table is straightforward. We could also find this quantity using the graphs of T R and T C.
Recall that, from the table, you determined the profit for a particular quantity by subtracting the value of
T C from the value of T R. Graphically, profit is the vertical distance between the graphs of T R and T C.
..... $
...↑|................T R
.
.
.
.
..
.
..
..
..... |P...........
..
.
..
.↓...... T C
.
.
..
.
.
.
.
.
..
.....
..
.
..
...........................................................................
.
..
..
...
.
...
..
...
..
..
..
q
10 Measure the vertical distance between the T R and T C graphs at q = 10. Compare this
distance to the value of profit at q = 10 in the table in exercise #2 and the profit graph in
exercise #8.
11 Using the TR and TC graphs, find the quantity q that gives the greatest profit. (This can be
hard on your eyes. I find that the best way to do a problem like this is to use a clear plastic
ruler and, holding it perpendicular to the horizontal axis, run it back and forth until you find
the place where the vertical distance between the graphs is greatest.) Check your answer
against your profit graph.
It seems that both you and your partner are partially right. It makes no sense to take an order of 13 reams,
because you lose money on it. The best size order is 6 reams, because you make the greatest profit on it. As
you take larger and larger orders between 6 and 12, you make less and less profit, but you still make a profit.
The odd thing is that your TR hits its peak at 8, but this number is not the significant one for profit.
Another way to find maximum profit is to consider marginal revenue (M R) and marginal cost (M C).
Marginal revenue is the change (positive or negative) in total revenue that comes with an increase of one
unit in sales.
For example, T R at q = 3 is $39, while T R at q = 4 is $48. In increasing sales from 3 reams to 4, we
increase T R by $9 ($48 − 39 = $9). We say that marginal revenue at q = 3 is $9. Using the ∆ notation
introduced in Worksheet #1, we’d say that marginal revenue is simply ∆T R, when ∆q = 1 ream. (Recall
that ∆ means “the change in.”)
12 Complete the M R row of the following table.
q
MR
MC
∆P
0
15
4
11
1
13
4
2
11
4
3
9
4
5
4
5
6
7
8
9
4
4
4
4
4
4
10
−5
4
−9
11
12
13
14
15
4
4
4
4
4
W ORKSHEET #3
Print Shop
31
The idea of marginal cost is similar: marginal cost is the change (positive or negative) in total cost that
comes with an increase of one unit in sales. (That is, M C = ∆T C when ∆q = 1.) Since our only cost is
the cost of paper, if we increase sales by one unit, then total cost goes up by the cost of one ream of paper.
That means that, for every quantity, marginal cost is $4.
We found M R at q = 3 to be $9 and M C at q = 3 is $4. That means, if a customer increases his order
from 3 reams to 4 reams, you’ll take in an aditional $9 and you’ll spend an additional $4. This means that
your profit will increase (by $5) when a customer increases his order from 3 reams to 4 reams.
13 In the table from exercise #12, fill in the last row. Under each value of q, indicate how the
profit will change if sales are increased to q + 1 reams. (Try to do this using only the values
of M R and M C in the table. Once you’ve completed the table, check your results against
the values of P in the table in exercise #2.)
14 Notice that q = 6 is the first quantity at which the M R drops below M C. Explain why this
implies that profit is maximized at q = 6.
We end this worksheet with a summary of some of the terms and procedures you’ve learned.
• Total revenue (T R) is the amount of money you receive from a customer for an order of a specific size.
Total cost (T C) is the amount of money you spend to fill an order. Profit (P ) is the difference between
T R and T C (P = T R − T C). Profit may be positive or negative.
• Given the graphs of T R and T C, the maximum possible profit occurs where T R is larger than T C
and you see the largest vertical distance between the two graphs.
• Marginal revenue is the additional total revenue you get with an increase in sales of one unit. (In this
worksheet, one unit is equal to one ream.) Marginal cost is the additional total cost you get with an
increase in sales of one unit. Profit is maximized at the first quantity at which M R falls below M C.
Worksheet #4
Increments and Speeds
Key Question
In Worksheet #1, we started with a graph of distance versus time and, in Worksheet #3, we drew
a graph of total revenue versus quantity. In the car example of Worksheet #1, we also defined
average speed (an incremental rate of change) and average trip speed (an overall rate of change).
What are the analogies of these speeds in the context of total revenue?
Recall the work you did with your ruler in Worksheet #1. You saw that one way to visualize average speed
over a five-minute interval was as the slope of a secant line. For example, on the interval from t = 25 to
t = 30 minutes, ∆D = 24.0 miles − 23.0 miles = 1.0 mile. The average speed over that interval is
1 mile
∆D
=
= 0.2 mpm.
∆t
5 minutes
This is the slope of the secant line through the points (25, 23) and (30, 24).
1
Complete the following table. Each entry in the last row is both an average speed over a fiveminute interval and the slope of a secant line to the distance graph. Note that the columns
are offset since they involve values from two adjacent columns.
for ∆D and ∆D
∆t
Slope
= ∆D
∆t
•
∆D
2
t
35
40
45
50
D
25.1
26.7
29.3
33.3
∆D
1.6
∆D
∆t
0.32
•
∆t
We’ll do the same in the context of revenue. Complete the following table.
•
∆T R
q
1
2
3
4
5
6
TR
15
28
39
48
55
60
∆T R
•
13
∆T R
∆q
32
∆q
Slope
R
= ∆T
∆q
W ORKSHEET #4
Increments and Speeds
33
If you did exercise #2 correctly, then you should see that the entries in the ∆T R-row are exactly the
R
same as the entries in the ∆T
∆q -row. (This is because ∆q = 1, of course.) Recall what we learned in
R
Worksheet #3: M R = ∆T R. So, we can say that M R is the same as ∆T R, ∆T R is the same as ∆T
∆q , and
∆T R
∆q can be thought of as the slope of a secant line through T R. Thus, we can think of marginal revenue as
the slope of a secant line through the graph of T R.
3
Go back to the graph you drew of T R in Worksheet #3, Exercise #3. Draw the secant line
through the T R graph at q = 4 and q = 5 and compute the slope of this line. Compare your
answer with your entries in the table in Worksheet #3, Exercsie #12. Did you just compute
M R at q = 4 or at q = 5?
When we have a variable y that depends on a variable x, then a change in x given by ∆x leads to a
∆y
change in y given by ∆y. The quotient ∆x
is called the incremental speed of y (with respect to x).
Thus the incremental speed over 5 minute intervals in the moving car is average speed, and incremental
speed over 1 unit intervals on the TR graph is MR.
We also have the overall speed of y (with respect to x), the change in y since the beginning divided by
the change in x since the beginning. If the graph of y goes through the origin, then overall speed is given by
the slope of a diagonal line and is calculated using the formula
overall speed =
y
.
x
4
If you take y to be D and take x to be t, then, looking at the graph at the start of Worksheet #1,
calculate the overall speeds for t = 5, 10, 15. Look back at Worksheet #1 to find out what
name we gave to the overall speed in the case of Distance vs. time.
5
If we start with the TR vs. q table (below) of the Print Shop, we can calculate the overall
speed for TR vs. q. We will call the overall speed in this case average revenue (AR), so
that the AR at q = 3 is 13. Again, the incremental speed in this case is MR. Complete the
following table:
q
TR
MR
AR
1
15
13
15
2
28
11
14
3
39
9
13
4
48
7
5
55
5
6
60
3
7
63
1
8
64
−1
9
63
−3
10
60
−5
11
55
−7
12
48
−9
6
Compare the values of average revenue to the values of price p in the table in Worksheet #3,
Exercise #2. What is the relationship between AR and p? Use the fact that AR = TqR and
T R = p × q (Worksheet #3, Exercise #2) to show mathematically that AR = p.
7
The following is a sketch of the TR vs. q graph. Recap the procedures for reading AR and
MR from this graph. Many students in economics courses confuse MR and AR. Describe,
in terms of this graph, how they are different.
34
W ORKSHEET #4
60
50
40
30
20
10
Increments and Speeds
...... T R
................................
....................
..........
.
.
..
.
.
.
.
.
.
.
.
........
....
.
..
.
.
.
.
.
........
.
....
..
.
.
.......
.
.
.
.
.
.
..
.
...
..
.
.
...
.
.
..
.
.
.
..
.
...
...
....
.
..
.
..
..
....
.
.
..
..
..
....
.
.
..
.. ......
.. .....
.. ...
.. ..
...................................................................................................................................................................................................................................
q
2
4
6
8
10
12
A Further Analogy
As a student, you get two kinds of measures of how you are doing in your courses. Each
quarter you get a GPA for that quarter (label it QGPA), which tells you how you have been
doing “recently,” and you get a cumulative GPA (CGPA) which tells you how you have been
doing overall.
As an example, let’s consider Terry’s grades during Terry’s first quarter at UW.
Course
Math 111
Psych 101
Engl 111
Music 116
8
Credits
5
5
5
2
Grade
3.7
3.3
2.6
4.0
Grade Points
18.5
16.5
To compute grade points for each course, we multiply the number of credits earned for that
course by the course grade. Complete the table above by computing the grade points Terry
earned for English and Music.
We compute Terry’s grade point average for the quarter (QGPA) by dividing the total number of grade
points for the quarter (QGP) by the total number of credits for the quarter (QCr).
9
Compute Terry’s QGP, QCr, and QGPA for this quarter.
The university also computes your cumulative grade point average (CGPA). Your cumulative grade point
total (CGP) is the sum of the grade points earned in every class you’ve ever taken (for a grade) at UW. Your
cumulative class credit total (CCr) is the total number of credits for every class you’ve ever taken (for a
grade) at UW. Your CGPA is your CGP divided by CCr.
W ORKSHEET #4
Increments and Speeds
35
10 The chart below gives the QGPA of a student who took 15 credits every quarter for 12
quarters. First complete the row labeled QGP, which gives the number of grade points the
student received during only that quarter. Then complete the CGP row, which gives the total
number of grade points the students has received up through that quarter. Finally, complete
the CGPA row, which gives the student’s CGPA as it stands at the end of each quarter.
CCr
QGP
QGPA
CGP
CGPA
15
60.0
4.0
60.0
4.0
30
52.5
3.5
112.5
3.75
45
39.0
2.6
151.5
3.37
60
75
90
105
120
1.9
1.4
1.1
217.5
2.42
1.0
1.1
249.0
2.075
135
21.0
1.4
150
28.5
1.9
165
39.0
2.6
180
3.5
390.0
2.17
11 On the axes below, sketch Cumulative Grade Points (CGP) vs. total credits (CCr). (Connect
the dots with a nice smooth curve.)
.
......
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.............................................................................................................................................................................................................................................
12 After the fourth quarter at UW, this student has earned 60 total credits and has a cumulative
GPA of 3.0. Describe how you could compute this CGPA using the graph of CGP above and
the slope of a line. Is the line you described a secant line or a diagonal line (or both)?
13 During the fourth quarter at UW, this student earns 15 credits and a quarterly GPA of 1.9.
Describe how you could compute this QGPA using the graph of CGP above and the slope of
a line. Is the line you described a secant line or a diagonal line (or both)?
14 The Key Question in Worksheet #1 was: If you are traveling along a freeway, is it possible to
have been increasing your speedometer readings recently while still decreasing your average
trip speed? This question now translates into: Is it possible to have increased your QGPA
recently while still lowering your CGPA? Can you answer this from your intuition about
grade point average? Answer this question by using a clear plastic ruler on the CGP graph
you drew in exercise #11.
36
→
W ORKSHEET #4
Increments and Speeds
15 A counter at the gate of a parking lot counts the number of cars that have entered the lot by
any given time. Another counter counts the number of cars that have exited the lot by any
given time. The two graphs below show the number of cars that have come in and gone out
over a 12-hour period. Suppose the lot had 1,000 cars in it at time t = 0.
4,000
a) Find the overall average rate of flow
into the parking lot over the first 7
hours.
OUT
3,000
IN
2,500
Cars
b) Name a 1-hour interval over which
at least 500 cars left the lot.
3,500
2,000
1,500
IN
1,000
c) How many cars are in the lot at 5
hours? How many cars are in the
lot at 10 hours?
OUT
500
1
2
3
4
5
6
7
8
9
10
11
12
t
HOURS
d) What is the incremental rate of flow out of the lot over the time period from 3 hours to 6
hours?
e) Name the time at which the lot will have the greatest number of cars in it. Name the time
when the lot will have the smallest number of cars in it. How many cars will be in the lot at
each of those times?
→
16 You sell Morkles. The following chart gives values of marginal revenue and marginal cost.
q (in Morkles)
M R (in dollars)
M C (in dollars)
0
20
10
1
18.61
7.81
2
17.22
6.25
3
15.83
5.31
4
14.44
5
5
13.06
5.31
6
11.67
6.25
a) What is your total revenue if you sell 6 Morkles?
b) What is your average revenue if you sell 3 Morkles?
c) Give the longest interval over which profit is increasing.
7
10.28
7.81
8
8.89
10
9
7.50
12.81
10
6.11
16.25
Functional Notation and Graphs
37
Worksheet #5
The Lagging Car
Introduction
A red car and a purple car travel along a long, straight road. At time t = 0, the two cars are at the
same location (we’ll call this the starting line). Information about Red’s distance, in miles from
the starting line, is given in the following table.
Time
Red
Purple
D(t)
5
8.5
0
10
14.6
8.5
15
18.7
20
21.2
25
23.0
30
24.0
35
25.1
40
26.7
45
29.3
50
33.3
55
39.4
60
48.0
Purple stays at the starting line for five minutes and then:
(∗)
For every place along the highway, Purple reaches that
place exactly 5 minutes after Red.
Key Questions
I. Is Purple always the same distance behind Red?
II. If not, when is Purple getting further behind and when is Purple getting closer?
1
Red is 8.5 miles from the starting line at t = 5 minutes. What does this mean about the
Purple car? Red reaches 14.6 miles at t = 10 minutes. What does this mean about the
Purple car? Fill in Purple’s distances in the table.
2
If you did exercise #1 correctly, then you should see that all of Purple’s distances are Red’s
distances shifted right one box.
(a) If Purple had a fifteen-minute delay, instead of five, how would Purple’s distances relate
to Red’s?
(b) If Purple was five minutes ahead of Red, instead of behind, how would Purple’s distances relate to Red’s?
39
40
3
W ORKSHEET #5
The Lagging Car
Here is the graph of Red’s distance from the starting line at time t.
.....
48 ....
...
.
.
...
.
..
R...
.
40 .
..
...
.
.
.
..
.
.
....
.
.
.
.
32 ...
.
.
.......
..
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.......................
.
.
.
.
.
.
.
.
24 ...
.
.
.
.
.
.
.
.
.
.
.
..
...........
.
.
.
..
.
.
.
.
.
16 ...
.....
.
.
.
.
..
....
..
.
.
.
.
8 .. ...
.. ..
.. ..
........................................................................................................................................................................................................................................................
5 10 15 20 25 30 35 40 45 50 55 60
(a) On the same set of axes, sketch Purple’s distance from the starting line at time t.
(b) How does the graph of Purple’s distance relate to the graph of Red’s?
(c) If Purple had a fifteen-minute delay, instead of five, how would Purple’s distance graph
relate to Red’s?
(d) If Purple was five minutes ahead of Red, instead of behind, how would Purple’s distance
graph relate to Red’s?
4
The distance between the two cars can be seen in the graphs in the same way we read profit
from the graphs of T R and T C in Worksheet #3: as the vertical distance between the two
graphs. Starting at t = 5, hold your ruler perpendicular to the time axis and slide the ruler to
the right, keeping it vertical. Which of the following describes the distance between the two
cars from t = 5 to t = 60?
(a) The distance between the two cars always increases.
(b) The distance between the two cars always decreases.
(c) The distance between the two cars decreases for a while and then increases.
(d) The distance between the two cars increases for a while and then decreases.
5
When are the two cars closest together?
6
Answer both of the Key Questions.
Mathematicians have short-hand notation for phrases like “the red car’s distance from the starting line
at time t” and “for every place along the highway, Purple reaches that place exactly 5 minutes after Red.”
W ORKSHEET #5
The Lagging Car
41
Let R(t) represent the red car’s distance from the starting line at time t. Then, for example, R(5) = 8.5
and R(10) = 14.6. The English translation of this last statement is: “The red car travels 14.6 miles in the
first 10 minutes.”
7
(a) Translate the following statement into English: R(15) = 18.7.
(b) Translate the following statement into functional notation: The red car travels 23.0 miles
in the first 25 minutes.
8
Let f (q) represent the total revenue for printing q reams of paper in the print shop example
of Worksheet #3.
(a) Translate into English: f (2) = 28.
(b) Translate into functional notation: The total revenue for printing 4 reams is $48.
The power of functional notation comes from our ability to combine the symbol R(t) (which stands here
for the distance covered by the red car in the first t minutes) with the operations of arithmetic. Thus R(15)−
R(10) is translated into “the distance covered by the red car in the first 15 minutes minus the distance
covered by the red car in the first 10 minutes.” The same thing could be said more concisely as “the distance
covered by the red car in the time from 10 to 15 minutes.”
9
Translate into comprehensible English the following statements. (Note that “<” is read “is
less than,” and “>” is read “is greater than.”) After you’ve translated, decide whether each
statement is true or false.
a) R(45) − R(30) = 2.7
b) R(30) > R(25) c) R(30) − R(25) < R(20) − R(15)
d) R(45 + 5) − R(45) = 2 e) R(20)
= 1.06
20
10 Translate the following statements into functional notation (involving R( )). Then decide
whether each statement is true or false.
(a) The red car traveled 3.7 miles in the time period from 25 to 40 minutes.
(b) The red car traveled further in the first 20 minutes than in the first 15 minutes.
(c) The red car traveled less in the time period from 35 to 45 minutes than in the time
period from 50 to 60 minutes.
(d) The distance that the red car traveled in the 10 minutes after the first 20 minutes was
4.3 miles.
(e) The red car’s average trip speed after 30 minutes is 0.8 mpm.
Now that we’ve addressed the notation, let’s take another look at the Key Questions. We will continue
to let R(t) be an abbreviation for the distance covered by the red car in the first t minutes. We’ll also let
P (t) represent the distance covered by the purple car in the first t minutes and we’ll let D(t) represent the
distance between the two cars at time t. In notational form, D(t) = R(t) − P (t).
42
W ORKSHEET #5
The Lagging Car
11 Go back to the table at the beginning of this worksheet and calculate the values for the
D(t)-row. The values should confirm your answers to the Key Questions that you obtained
graphically.
Now, if t is any particular time, then t − 5 is five minutes earlier. (Do you believe that?) At any particular
time t, the purple car is where the red car was five minutes earlier. We can express this in functional notation
as follows:
P (t) = R(t − 5).
So we can rewrite the formula for D(t):
D(t) = R(t) − P (t) = R(t) − R(t − 5).
But R(t) − R(t − 5) is the distance Red has traveled from time t − 5 to time t. (Do you believe that?) We
can translate the statement D(t) = R(t) − R(t − 5) into English as: “The distance between the cars at time
t is the same as the distance the red car traveled over the preceding five minutes.”
12 Express D(20) as the difference of two values of R.
13 Suppose you could change the graph of Red’s distance so that the purple car caught up to the
red car after 20 minutes. Use your answer to exercise #12 to explain what the red car would
have to do in order for this to happen.
14 A counter at the gate of a parking lot keeps track of the number of cars that have come into
the lot since noon. Another counter keeps track of the number of cars that have left the lot
since noon. The two following graphs show the number that have come in and gone out over
a 12 hour period. Let C(t) represent the number of cars in the lot t hours past noon. Suppose
the lot has 1,000 cars in it at noon (C(0) = 1000).
a) Translate the following into
English and then decide if
the statement is true or false:
C(9) > C(6).
4,000
3,500
OUT
3,000
IN
2,500
b) The In and Out graphs cross at
t = 7.6. What is the value of
C(7.6)?
Cars
→
2,000
1,500
IN
1,000
c) Name the longest time interval over which the overall rate
at which cars come into the lot
is declining.
OUT
500
1
2
3
4
5
6
7
8
9
10
11
12
t
HOURS
d) Suppose the flow out of the lot is changed so that every car leaves the lot exactly 2 hours
later than shown on the graph. Under this new assumption, what is the value of C(6)?
Worksheet #6
A Reservoir in Three Languages
Key Question
Amount (thousands of gallons)
The graph below gives amounts of water that have come into a reservoir over a 12 hour period.
(That is, the reading at each time tells how much water has flowed into the reservoir from the
beginning of the 12 hour period to that time).
.....
30 ...
..
...................
.
.
.
27 ...
.
.
.
.
.
.
...
......
.
.
.
24 ...
.
..
....
.
.
.
21 ..
..
.
..
.
18 ...
.. ..
..
.
15 ...
.. ..
.
..
12 ...
..
.
.
..
..
9 ...
.
.
..
..
.
6 ...
.
.
....
..
.
.
.
.
.
.
3 .
..
.........
.......................................................................................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
Express each of the following in functional notation and tell how you would use the graph to arrive
at an answer. (We will wait until Worksheet #7 to answer these questions. For now, just translate.)
A. Find a time such that the amount of water that flows into the reservoir from t = 2 hours to
that time is 16 thousand gallons.
B. How much time would you need to wait beyond 5 hours to have an additional 18 thousand
gallons come into the reservoir?
C. Find the time at which the (overall) average rate of flow of water into the reservoir from the
beginning (midnight) is 2 thousand gallons per hour.
D. Find a time interval of length 12 -hour over which the (incremental) rate of flow over that time
interval is 4 thousand gallons per hour.
43
44
1
W ORKSHEET #6
A Reservoir in Three Languages
Using the graph, tell what each of the following quantities is equal to:
(a) the amount of water that has flowed into the reservoir from 2 hours to 8 hours;
(b) the amount of water that has flowed into the reservoir from time t = 5 hours to a time
3 hours later;
(c) the (overall) average rate of flow of water into the reservoir from midnight to time t = 9
hours;
(d) the rate of flow of water into the reservoir over the half-hour interval starting at t = 3
hours.
The wording in these questions and in the Key Questions is very cumbersome. Functional Notation
is the way to compress the wording and be able to answer the Key Questions.
Let A(t) stand for the amount of water that has come into the reservoir since midnight at a time t
hours after midnight.
2
The following are translations of the items in exercise #1(a) through (d). Tell which is which.
A(9)
9
ii) A(8) − A(2)
i)
iii) A(5 + 3) − A(5)
iv)
3
A(3 + 12 ) − A(3)
1
2
Using the graph, find the following quantities, and then tell what these quantities are in terms
of the water (e.g. “the amount of water that has . . . ”)
a) A(10) − A(3)
A(5)
5
A(6 + 2) − A(6)
c)
2
b)
In order to answer the Key Questions we need a more sophisticated use of functional notation. We
need to be able to translate very easily between quantities “at the reservoir,” quantities described in
functional notation and quantities pictured on the graph.
W ORKSHEET #6
4
A Reservoir in Three Languages
45
In the following chart are listed four quantities “at the reservoir.” Each Key Question is about
one of these quantities. In the box next to each quantity put the letter (A, B, C, D) belonging
to the Key Question that is about this quantity. FOR NOW, DON’T FILL IN ANY MORE
OF THE CHART.
Reservoir
Functional Notation Graph Language
The (overall) average rate
of flow from midnight to
time t
The amount of water that
flows in from 2 hours to t
hours
The amount of water that
flows in from 5 hours to a
time h hours later
The (incremental) rate of
flow from a time t hours
to a time 21 -hour later
Remember that A(t) stands for the amount of water that has flowed into the reservoir by time t.
Thus A(7) − A(4) is the number that results from subtracting A(4) from A(7), and is equal to the
amount of water that flowed in between 4 AM and 7 AM. Likewise, A(6)
6 is the number that results
from dividing the amount of water that has come in by 6 AM by 6 hours, and is equal to the (overall)
average rate of flow from Midnight to 6 AM.
5
Each of the following quantities (in functional notation) is a quantity in one of the Key
Questions. Determine the English translation of each and place it in the appropriate box in
the Functional Notation column of the table in exercise #4.
i) A(5 + h) − A(5)
ii)
iii)
A(t)
t
A(t + 12 ) − A(t)
1
2
iv) A(t) − A(2)
This chart is now becoming a dictionary for translating between the three languages: reservoir, functional
notation and graph. By using the first two columns of the dictionary, check that the following are correct
translations of the Key Questions.
46
W ORKSHEET #6
A Reservoir in Three Languages
Revised Key Questions
A. Find a value of t such that A(t) − A(2) = 16.
B. Find a value of h such that A(5 + h) − A(5) = 18.
C. Find a value of t such that
A(t)
= 2.
t
D. Find a value of t such that
A(t + 21 ) − A(t)
1
2
= 4.
In order to actually solve the Key Questions, we need to be able to read these quantities from the graph.
That is, we need to fill in the third column of the dictionary.
6
The following four quantities can be read from the graph. Put each one in the appropriate
box in the Graph Language column of the table in exercise #4.
i) The slope of the diagonal to the place on the graph at t hours.
ii) The change in the height of the graph from 2 hours to t hours.
iii) The slope of the secant from t hours to t +
1
2
hours.
iv) The change in the height of the graph from 5 hours to a time h hours later.
Now that you have the third column of the dictionary, check that the following are correct translations of
the Key Questions.
Second Revised Key Questions
A. Find a value of t such that the change in the height of the graph from 2 to t is 16.
B. Find a value of h such that the change in the height of the graph from 5 to 5 + h is 18.
C. Find a value of t such that the slope of the diagonal to t is 2.
D. Find a value of t such that the slope of the secant from t to t +
1
2
is 4.
We will answer the Key Questions in Worksheet #7. We end this worksheet with one more exercise in
which you’ll practice your translation skills.
W ORKSHEET #6
7
A Reservoir in Three Languages
47
Complete the following chart with equivalent phrases in the indicated “language.”
English
Functional Notation
Graph Language
A(4) − A(1)
the overall average rate
of flow after 5 hours
the slope of the secant
line from t = 1 to
t=1+h
the change in height
from t = 1 to t = T
A(4)
4
the incremental rate of
flow from t = 3 to a
time h hours later
(There is one more problem that begins on the next page.)
48
→
8
W ORKSHEET #6
A Reservoir in Three Languages
The graphs of f (x) and g(x) are given below.
y
16
14
y = f (x)
12
10
8
6
4
y = g(x)
2
x
1
2
3
4
5
6
7
8
9
10
(a) Find each of the following quantities from the graphs:
f (6) − f (1)
f (3.05) − f (3.0)
i)
ii)
iii) f (7) − g(7)
5
0.05
(b) For each of the statements below, find an interval of length 2 for which the statement is
true.
i) g(x) decreases and then increases.
f (x)
increases.
ii)
x
iii) f (x + 1) − f (x) is positive, then 0, then negative.
(c) For each of the following statements find the desired value(s) of x.
i) Find an x such that f (x) ≥ 6 and f (x) − g(x) ≥ 3.
f (x)
is decreasing.
x
iii) Find the largest value of f (x) in the interval over which g(x) is decreasing.
ii) Find an interval of x such that f (x) is increasing, but
W ORKSHEET #6
A Reservoir in Three Languages
49
(d) The graphs A–F below are common shapes of graphs. For each of the quantities listed
below, tell which of the shapes of graph would be closest to the graph of that quantity
over the given interval.
(A)
(B)
(C)
(D)
(E)
(F)
f (x + 0.2) − f (x)
from 2 to 4.
0.2
f (x)
from 5 to 9.
ii)
x
iii) f (x + 0.3) − f (x) from 5 to 7.
i)
Worksheet #7
Increments and Reference Lines
Key Questions
Amount (thousands of gallons)
The graph below gives amounts of water that have come into a reservoir over a 12 hour period.
......
30 ...
...
....................
.
.
27 ...
.
.
.
.
.
.
.
.
......
.
.
24 ...
.
.
..
...
.
.
.
.
21 .
..
..
.
.
.
18 ..
..
.
..
.
15 ...
.. ..
..
.
..
12 ...
.
..
.
..
.
9 ...
.
..
..
.
.
.
.
.
6 .
..
....
.
.
.
.
.
3 ..
......
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.................................................................................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
Answer the Key Questions from Worksheet #6.
A. Find a time such that the amount of water that flows into the reservoir from t = 2 hours to
that time is 16 thousand gallons.
B. How much time would you need to wait beyond 5 hours to have an additional 18 thousand
gallons come into the reservoir?
C. Find the time at which the (overall) average rate of flow of water into the reservoir from the
beginning (midnight) is 2 thousand gallons per hour.
D. Find a time interval of length 12 -hour over which the (incremental) rate of flow over that time
interval is 4 thousand gallons per hour.
50
W ORKSHEET #7
Increments and Reference Lines
51
1
Remind yourself of the functional notation and graphical translations of Key Question A
from Worksheet #6 and answer Key Question A.
2
The technique you use to answer Key Question B should be very similar to the technique
you used to answer Key Question A, but B requires one additional step. Remind yourself of
the functional notation and graphical translations of Key Question B, answer the question,
and explain why you need this extra step.
Remind yourself of the functional notation and graphical translations of Key Question C. To solve this
problem, you need a diagonal line with slope 2. Since diagonal lines go through the origin, you’ll want to
2
draw a line through the origin with slope 2. So we have: slope = rise
run = 1 . You want to draw the line that
goes through the origin and through the point that is 2 units up and 1 unit to the right of the origin. However,
the marks on the vertical axis are in multiples of 3, which makes a “rise” of 2 units difficult to find with any
accuracy. We can express the slope as a fraction that is equivalent to 12 , whose numerator is a multiple of 3:
slope =
rise
2 3
6
= · = .
run
1 3
3
The diagonal line with slope 2 is the line that goes through the origin and the point 6 units up from and
3 units to the right of the origin.
3
Draw the diagonal line with slope 2 and use it to answer Key Question C.
Remind yourself of the functional notation and graphical translations of Key Question D. To answer this
question, we need a secant line whose slope is 4. It will help us if we know what a line with slope 4 looks
4
like. Again, slope = rise
run = 1 . But the units on the vertical axis make it difficult to see a “rise” of 4.
4
Express 14 as an equivalent fraction whose numerator is a multiple of 3. Draw any line on
the reservoir graph that has slope 4. This will be your reference line.
Two lines are parallel if they have the same slope. To answer Key Question D, we need a secant line
with slope 4 and we have just drawn a line with slope 4. In particular, we need to find a line, parallel to
our reference line, that goes through the reservoir graph at two points that are 12 -hour apart. We can use
another version of the “rolling ruler method” to accomplish this.
5
Place your ruler along your reference line. Slide your ruler, either toward you or away from
you, keeping it parallel to your reference line, until it goes through two points on the reservoir
graph that are exactly 12 -hour apart. Answer Key Question D.
Reference lines are handy when you want to find a secant line with a specified slope. But they can also
sometimes come in handy when the question doesn’t seem to involve a slope.
Consider a new Key Question:
E. Find a 1.5-hour interval during which 6 thousand gallons flow into the reservoir.
6
Translate Key Question E into functional notation and into graphical language.
52
W ORKSHEET #7
Increments and Reference Lines
Your graphical language translation should be something like: “Find a value of t such that the change
in height from t to t + 1.5 is 6.” So, this does not appear to be a question about slopes. As stated, we need
to find two points on the reservoir graph that are 1.5 hours apart and between which the height changes by
6 thousand gallons. This is very difficult. BUT we can turn this into a question about slopes to make things
a bit easier. The “run” between these two points is 1.5 and the “rise” is 6. So, the slope of the line between
6
or, simply, 4. The following question will give us the same answer as Key Question E, and is
them is 1.5
easier to answer:
E’. Find a value of t such that the secant line from t to t + 1.5 has slope 4.
This is very similar to Key Question D.
Use the reference line you drew in exercise #4 and the “rolling ruler method” to answer Key
Question E’, thereby answering Key Question E.
8
Below are the graphs of weight in pounds of two puppies, A and B (from the same litter) for
a period of 72 hours after their birth. We denote the weight of the two puppies at time t by
A(t) and B(t), respectively.
.....
3.2 ..
...............
.
.
.
.
..
.
.
.
.
B
.
.
.......
2.8 ...
.
.
.
.
..
....
.
.
2.4 ..............
.
...
.. .........
.
.
.
.
.
.
.
.
...........
2.0 ......
....
............
.
.. ...
.
.
.
............ .......
...
. ..
.
.
.
1.6 ... .......
.
.
.
.
.....................
...........
......A
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..............
...............
1.2 ..
..
.
0.8 ...
..
0.4 ...
..
..........................................................................................................................................................................................................................
t
6 12 18 24 30 36 42 48 54 60 66 72
Hours
Pounds
→
7
a) What is the difference in the weights of the two puppies at 42 hours?
b) Name a 6-hour time interval over which B has gained 0.4 pounds.
) is 0.01. (Note: Since
c) Find a time when B’s overall average weight gain (i.e. B(t)−2.00
t
B’s initial weight is not zero, the graph of B(t) does not go through the origin. Thus,
the overall rate of change is the slope of a secant line that is not a diagonal line.)
d) Find the incremental rate of change of B’s weight over the 6-hour time period starting
at t = 54 hours.
W ORKSHEET #7
Increments and Reference Lines
53
e) What is the greatest difference in the two puppies’ weights over the 72 hours?
f) Suppose a third puppy, Puppy C is born in the same litter and always weighed exactly
3
of A’s weight. How much weight did Puppy C lose over the first 24 hours?
4
→
9
The graph to the right shows the
price per share (in dollars) of the
common stock of Technigraphics
(TG) over a 12-month period.
70
60
50
TG
40
a) Suppose another stock, SureThing Inc. (ST) starts at $10
per share and increases its
price every month by $5. On
the axes to the right draw the
graph of the price of ST for
the 12-month period.
30
20
10
t
0
1
2
3
4
5
6
7
8
9 10 11 12
(Months)
b) Find the change in the prices of each of the stocks in the period 1 month to 3 months.
c) Give a time when the incremental rates of change of the two stocks from that time to
one month later are the same.
d) Let f (t) be the abbreviation for the price of TG at time t. Find the value of
f (6) − f (0)
.
6
e) Name a 2-month time interval, if any exists, over which the incremental rate of change
f (t + 0.1) − f (t)
changes from positive, to zero, to negative.
0.1
f) Find a 2-month period over which the price of TG increases by $15.
g) Give the time at which the difference in the prices of the stocks is greatest.
h) Which stock has the larger change in price from the beginning of the period to the time
t = 5 months?
→
W ORKSHEET #7
10 The graph to the right is of the
amount of water that has flowed
into a reservoir by various times
over a 12-hour interval starting
at Midnight. We abbreviate this
amount by I(t).
a) Water flows out at a constant
rate of 2,000 gallons per hour.
On the axes to the right draw
the graph of Water Out vs.
time.
b) What is the smallest amount
of water we can start with to
make sure there is always water available in the reservoir?
30
Amount (thousands of gallons)
54
27
24
21
18
15
12
9
6
3
Increments and Reference Lines
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1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
the “overall average rate of flow in.” Find the time at which
c) We will call the rate I(t)
t
this rate is greatest.
d) Suppose the reservoir has 9,000 gallons of water in it at Midnight. At what time will it
have 12,000 gallons of water in it?
e) Let g = I(t) and ∆g = I(t + 1) − I(t). Find two times at which ∆g = 2, 000.
f) Find the longest time interval starting at t = 6 hours during which the level of water is
rising.
I(t)
I(t + 0.25) − I(t)
<
.
g) Find a time t greater than 4 when
0.25
t
→
11 The graphs of f (x) and g(x) are given below.
y
16
14
y = f (x)
12
10
8
6
4
y = g(x)
2
x
1
2
3
4
5
6
7
8
9
10
W ORKSHEET #7
Increments and Reference Lines
For each of the following statements find the value of x that makes the equation true.
a)
g(x + 0.02) − g(x)
f (x + 3) − f (x)
= 2 b)
= 0 c)f (x + 2) − f (x) = 4
3
0.02
55
Worksheet #8
Analysis of Costs I
Introduction
Cost (Dollars)
A farmer produces potatoes that will be shipped to a processor that makes french fries for fast food
restaurants. The (heavy) graph below indicates the total cost (T C) of producing q bags of potatoes
for production levels between 0 and 700 bags. At any given time there is a market price of $p per
bag at which all potatoes are sold. (Note: In a market price situation, factors influencing the market
set the price, not the farmer. The price can fluctuate easily.)
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50 100 150 200 250 300 350 400 450 500 550 600 650 700
Quantity
Key Question
I. Determine the set of market prices that provide the farmer an opportunity to make a profit.
II. Determine the set of market prices that provide the farmer no opportunity to make a profit.
56
W ORKSHEET #8
1
Analysis of Costs I
57
Suppose the market price of a bag of potatoes is $0.75.
(a) Determine the total revenue (T R) for the production and sale of 100 bags if they sell at
this price. Do the same for 300 bags and again for 600 bags.
(b) The graph given at the beginning of this worksheet contained the graph of total cost
(T C). On the same set of axes, use the values you obtained in part (a) to sketch the
graph of total revenue (T R) for a market price of $0.75 per bag. The graph of T R
should be a straight line. What is its slope? Where does it intersect the vertical axis?
(c) Determine the production levels (quantities) that yield a profit.
2
Repeat exercise #1 with a market price of $0.35 per bag.
We’ve discovered that a market price of $0.75 per bag provides an opportunity to make a profit, depending
on the production level. A market price of $0.35 provides no opportunity to make a profit and the farmer
would most likely decide not to dig up the potatoes and ship them to the processor. We’ve also found that,
if each bag is sold at a price of $p, then the graph of T R is a line with slope p through the origin. So what
should the farmer do if the market price is $0.60 per bag?
3
Does the market price of $0.60 per bag provide the farmer an opportunity to make a profit?
If yes, then determine the production levels that make a profit.
So, you should now know how to tell whether a given market price will allow the farmer to make a profit.
But we don’t want to keep drawing lines every time the price changes. There should be a special market
price that acts as a transitional price. If the market price is lower than that special price, then the farmer
cannot possibly make a profit. If the market price is higher than that special price, then the farmer will know
that a profit is possible. We will call this special market price the breakeven price.
4
Place your ruler so that it lays along the q-axis. Keeping one end on the origin, increase
the steepness of your ruler until it first intersects the T C curve. Draw this diagonal line. It
should intersect the T C curve only once.
The diagonal line that you just drew is the least steep line that intersects the T C graph. This line is a line
of transition. Any line steeper is above the T C graph for some quanitities and a profit is possible. Any line
less steep is always below the T C graph and no profit is possible.
5
Compute the slope of the diagonal line you drew in exercise #4. This slope is the breakeven
price for this T C graph.
6
Answer the Key Questions.
Now notice that the height of the T C graph at q = 0 is 80. That is, even if the farmer produces no
potatoes, the farmer must pay out $80 in costs. The value of T C when q = 0 is called the fixed cost (F C).
(Usually, fixed cost includes costs like space and equipment rental, loan payments, some salaries, and so
58
W ORKSHEET #8
Analysis of Costs I
forth.) Total cost is made up of fixed cost plus variable cost (V C). Variable cost at q is the amount it costs
the farmer to produce q bags of potatoes without including the fixed cost. That is, at each quantity q,
T C(q) = V C(q) + F C or V C(q) = T C(q) − F C.
The graph of V C is given at the beginning of this worksheet. Notice that, at any value of q, the vertical
distance between the T C and V C graphs is always 80 (the value of F C). Also notice that the graphs of
T C and V C have exactly the same shape: T C is just V C shifted up by the amount of F C.
7
Suppose the market price for a bag of potatoes is $0.35. You drew the T R graph for this
market price in exercise #2.
(a) Suppose the farmer produced 200 bags of potatoes and sold them at this price. How
much money would the farmer lose? How does this compare to the farmer’s fixed cost?
(b) Suppose instead that the farmer produced 450 bags and sold them at this price. How
much would the farmer lose? How does this compare to the fixed cost?
(c) Find the set of quantities for which the farmer’s loss will be less than the fixed cost
at this market price. (Hint: The vertical distance between the T C and V C graphs is
always equal to F C.)
8
Suppose the market price is $0.15 per bag. Are there any quantities at which the farmer’s
loss is less than the fixed cost?
If the market price is $0.15 per bag, then the farmer is better off shutting down (and paying out only the
fixed cost of $80) than producing any amount of potatoes. On the other hand, if the market price is $0.35 per
bag, even though the farmer can’t make a profit, there are some quantities that will allow the farmer to make
back some of the fixed cost. Again, there should be a special transitional price. If the market price is lower
than that special price, then the farmer will be better off shutting down and paying out only the fixed cost. If
the market price is higher than that special price, then the farmer will know that it is possible to make back
at least some (if not all) of the fixed cost. We will call this special transitional price the shutdown price.
9
Place your ruler so that it lays along the q-axis. Keeping one end on the origin, increase
the steepness of your ruler until it first intersects the V C curve. Draw this diagonal line. It
should intersect the V C curve at the origin and at only one other point.
10 Compute the slope of the diagonal line you drew in exercise #9. This slope is the shutdown
price for this T C graph.
Notice that the procedure for finding the shutdown price was the same as the procedure for finding the
breakeven price — the only difference being that finding the shutdown price uses the V C graph and finding
the breakeven price uses the T C graph.
We end this worksheet with another way to view the breakeven and shutdown prices.
If the farmer produces q bags of potatoes, then the average cost (AC) per bag is given by the formula
AC(q) =
T C(q)
.
q
W ORKSHEET #8
Analysis of Costs I
59
(Note the functional notation. AC(q) is the average cost for producing q bags. T C(q) is the total cost for
producing q bags.)
11 Compute the average cost of producing 300 bags of potatoes. Notice that average cost is the
same as the slope of the diagonal line through the T C graph at q = 300.
12 Using a clear plastic ruler to indicate the slope of the diagonal line, describe the way the
values of AC vary as q goes from 100 to 700. In other words, do they increase the whole
way, decrease the whole way, or go one way and then the other? (This should be familiar to
you. It is identical to what you did in reading average trip speed from the distance graph in
Worksheet #1.)
13 Again using your ruler, find the smallest value that AC reaches. Notice that this is the
breakeven price you computed in exercise #5.
Does it make sense that the smallest value of AC is the same as the breakeven price? The following is a
mathematical justification of this fact.
TC
We defined AC to be
. This means that AC × q = T C. Also, recall that T R is price per item (our
q
market price p) times quantity. That is, p × q = T R.
If the market price is less than the smallest AC, then p < AC for every quantity. This means that
p × q < AC × q for every quantity q. Thus, T R < T C for every quantity q and no profit is possible.
On the other hand, if the market price is greater than the smallest AC, then p > AC for some quantities.
If q is one of those quantities, then p × q > AC × q and, therefore, T R > T C at that quantity and profit is
possible. So, the smallest AC is the transition from no possibility for profit to some possibility for profit.
We can do the same sort of thing with variable cost and shutdown price. If the farmer produces q bags of
potatoes, then the average variable cost (AV C) per bag is given by the formula
AV C(q) =
V C(q)
.
q
(Again, notice the functional notation.) AV C(q) is equal to the slope of the diagonal line through the V C
graph at q.
14 Using your ruler, find the smallest value of AV C. Check that this is the same as the shutdown
price you computed in exercise #10.
You learned several new terms and procedures in this worksheet. Here is a summary.
• Total cost is made up of two components: fixed cost and variable cost.
T C(q) = F C + V C(q).
• The breakeven price is a transitional price.
– If the market price p is less than the breakeven price, then no profit is possible regardless of the
quantity sold.
60
W ORKSHEET #8
Analysis of Costs I
– If the market price p is greater than the breakeven price, then there are some quantities that will
yield a profit.
• The breakeven price is the slope of the least steep diagonal line that intersects the T C graph.
• The shutdown price is also a transitional price.
– If p is less than the shutdown price, then you will always lose more than your fixed cost regardless
of the quantity sold.
– If p is greater than the shutdown price, then there are some quantities that will allow you to make
back some (if not all) of your fixed cost.
• The shutdown price is the slope of the least steep diagonal line that intersects the V C graph.
• Average cost and average variable cost are defined by the following formulas:
AC(q) =
T C(q)
V C(q)
and AV C(q) =
.
q
q
• The breakeven price is the smallest possible value of AC.
• The shutdown price is the smallest possible value of AV C.
→
15 The graphs which follow are of Total Revenue (TR) and Variable Cost (VC) for producing
Blivets. The Fixed Costs (FC) of producing Blivets is $200.
a) Find the Average Variable Cost
and Average Cost of producing 50
Blivets.
b) What is the Marginal Revenue of
manufacturing 30 Blivets?
c) Find the value of q at which the
profit of manufacturing Blivets is
greatest.
d) Recall that there are Fixed Costs.
What is the greatest profit that
can be realized from manufacturing Blivets?
1
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e) Give an interval of values of q, if any exists, over which MR and MC are both increasing.
f) What is the largest value of q at which any profit can be made?
g) Find a value of q greater than q = 30 for which AR = AVC.
Worksheet #9
Analysis of Costs II
Key Question
The set-up is exactly the same as in the previous worksheet. The graphs of TC and VC vs. q are
given again below. Now that you have found the breakeven and shutdown prices, assume that the
market price for bags of potatoes is $0.75 each. What quantity should you produce in order to
maximize your profit?
480
440
400
360
Cost (Dollars)
320
280
240
200
160
120
80
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50 100 150 200 250 300 350 400 450 500 550 600 650 700
Quantity
Recall from Worksheet #3, that the formula for profit is P = T R − T C. Graphically, profit is the vertical
distance between the T R graph and the T C graph. Thus, finding the quantity that produces the maximum
profit translates to finding the quantity that gives the greatest vertical distance between the T R and T C
graphs. (When T R is greater than T C, of course.)
1
(a) Draw the T R graph if the market price is $0.75 per bag. (Remember that, when the
price per item is the same for all quantities, T R will be a straight line through the origin
61
62
W ORKSHEET #9
Analysis of Costs II
whose slope is the market price. This is different than our Print Shop example since
there, the price per item changed depending on the quantity sold.)
(b) Use your ruler to find the quantity that leads to the greatest profit at a market price of
$0.75 per bag. Hold the ruler vertically and move it left and right to find the greatest
vertical distance between the two graphs.
(c) What is the maximum possible profit if market price is $0.75 per bag?
2
Suppose someone else is paying all your fixed cost and they don’t expect you to repay them.
Then, to determine profit, you look at the vertical distance between the T R and V C graphs.
(Note: This is a very unusual situation that applies to this exercise only.)
(a) At a market price of $0.75 per bag, what quantity leads to the largest profit?
(b) What is the largest profit if the market price is $0.75 per bag?
(c) The quantity that maximizes profit is the same whether you look at the total cost graph
or the variable cost graph. Explain why both graphs give the same quantity. (Hint:
Think about the vertical distance between the T C and V C graphs.)
3
Suppose the market price jumps to $1 per bag. Draw the new T R graph on the axes at
the beginning of the worksheet. Explain why finding the maximum profit by measuring the
vertical distance between T R and T C (or V C) would be troublesome in this case.
We can find the quantity that yields maximum profit if the market price is $1 per bag; but we’ll need a
different method.
Recall the definition of marginal revenue given in Worksheet #3: M R at q is the change in T R as we
increase production by one item. That is,
M R(q) = T R(q + 1) − T R(q).
Using the techniques of Worksheet #7, we know we can think of this, not only as an incremental change in
T R, but also as the slope of a secant line through the T R graph.
4
Suppose the market price is $1 per bag.
(a) Compute M R at q = 100 bags, q = 300 bags, and q = 600 bags.
(b) Describe the secant line through any two points on the T R graph.
We define marginal cost similarly: M C at q is the change in T C that comes from increasing production
by one item. That is,
M C(q) = T C(q + 1) − T C(q).
Again, we can think of this as both an incremental change in T C and as the slope of a secant line throught
the T C graph.
However, it’s difficult to distinguish between q = 50 and q = 51, for example, on this graph. Rather
than trying to draw a proper secant line through T C at q = 50 and q = 51, you will actually get a better
estimate for the M C at q = 50 by computing the slope of the tangent line to T C at q = 50. The tangent
line at q = 50 looks a lot like the secant line through q = 50 and 51 — but the tangent line just skims the
graph of T C at q = 50.
W ORKSHEET #9
5
Analysis of Costs II
63
Use the secant/tangent line approach to compute M C at q = 100, q = 300, and q =
600 bags.
Now that we have both M R and M C, let’s review the economist’s argument from the end of Worksheet #3.
Since T R is linear, M R is always the same: the slope of the T R graph ($1 in our case). At q = 100 bags,
M C is $0.45, less than M R. If a customer increases his order from 100 bags to 101 bags, then the farmer
will take in an additional $1 in revenue and spend an additional $0.45 in costs. The farmer’s profit will
increase by $0.55. Profit increases whenever M R is larger than M C. On the other hand, at q = 650 bags,
M C is $1.33 (you should check that you believe that). M R is still $1. So, M R is smaller than M C. If a
customer increases his order from 650 bags to 651 bags, then the farmer will take in an additional $1 in
revenue and spend an additional $1.33 in costs. The farmer’s profit will decrease by $0.33. Profit decreases
whenever M R is smaller than M C. It makes sense then that profit will be highest when we change from
M R > M C to M R < M C. So, profit is maximized when M R = M C.
M R is the slope of the linear T R graph and M C is the slope of the tangent line to the T C
graph. Lay your ruler along the T R graph for a market price of $1. Keeping the angle of
your ruler the same, slide the ruler down until it is just tangent to the T C graph. The point
at which this tangent line intersects the T C graph is where M C = M R. Find the quantity
that yields maximum profit at this market price. Can you tell what the largest profit is?
We give one further method for finding the quantity that
maximizes profit. At right is the graph of marginal cost vs.
q. Use your responses to exercise #5 to understand how this
graph is constructed.
7
(Dollars per Bag)
6
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
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1
2
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5
6
7 q
(Hundreds of Bags)
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Suppose the market price is $0.80 per bag.
(a) Compute the marginal revenue at q = 100, q = 300, and q = 600 bags.
(b) On the axes above, sketch the graph of M R.
(c) Recall that profit is maximized when M R = M C (a transition from M R > M C to
M R < M C). Use the graphs of M R and M C to find the quantity that yields maximum
profit at this market price.
8
What quantity yields maximum profit if the market price is $0.90 per bag?
64
W ORKSHEET #9
Analysis of Costs II
Now we know three different methods of finding the quantity that yields maximum profit:
• find the place where there is the greatest vertical distance between T R and T C;
• find the place where T R and T C have the same “slope”;
• find the place where the M R and M C graphs intersect.
Since we’ve already got the graph of M C, let’s go through another way to compute the breakeven price
and the shutdown price of Worksheet #8.
9
Recall from Worksheet #8 the definitions of average cost (AC) and average variable cost
(AV C):
T C(q)
V C(q)
AC(q) =
and AV C(q) =
.
q
q
Use the graph of T C at the beginning of this worksheet to complete the following table with
appropriate values of AC and AV C.
q
AC
AV C
100
1.50
0.70
200
300
0.67
400
500
0.50
0.34
600
700
0.53
The graphs of AC and AV C are included on the graph we gave of M C. Check the results
in your table against that graph.
Recall from Worksheet #8 that the breakeven price is the smallest value of average cost. Notice that on
the graph of AC, the lowest point is where the graph of AC crosses the graph of M C.
10 Explain why the breakeven price is both a value of AC and a value of M C. (Your explanation should involve slopes, diagonal lines, and tangent lines.)
11 Look again at the graph of M C. We get a value of M C by computing the slope of a tangent
line to T C. Explain why we would get the same values if we computed the slope of a tangent
line to V C. (Thus, we can construct the graph of M C from the graph of V C.)
12 Explain why the shutdown price is both a value of AV C and a value of M C.
W ORKSHEET #9
Analysis of Costs II
65
Here are some important facts that you have learned that will be helpful as you solve exam-type problems.
• Maximum profit occurs when T R > T C and we see the greatest vertical distance between the T R
and T C graphs.
• Maximum profit occurs when T R > T C and the T R and T C graphs have “matching slopes.”
• Maximum profit occurs when the graphs of M R and M C intersect in transition from M R > M C to
M R < M C.
• The breakeven price is the “y”-coordinate of the point at which the graphs of M C and AC intersect.
• The shutdown price is the “y”-coordinate of the point at which the graphs of M C and AV C intersect.
→
13 You sell items. The following is the graph of total cost.
2500
2250
2000
1750
TC
1500
dollars 1250
1000
750
500
250
0
0
200
400
600
800
1000
quantity (in items)
1200
1400
(a) What is the value of fixed cost?
(b) Estimate the value of marginal cost (M C) at q = 400 items.
(c) What is the breakeven price?
(d) What is the average variable cost (AV C) at q = 300 items?
(e) Items sell for $2.50 each. Name the smallest quantity at which T R = T C.
66
→
W ORKSHEET #9
Analysis of Costs II
14 You are manufacturing and selling artificial thumbs. The following graph shows the marginal
cost (MC) and average variable cost (AVC) of manufacturing various quantities of thumbs.
9
8.5
8
7.5
7
6.5
6
5.5
dollars
5
per thumb4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
MC
AVC
0
0.5
1
1.5
2
2.5
3
3.5
hundreds of artificial thumbs
4
4.5
5
5.5
6
Use the graph to answer the following:
(a) Find the Shutdown Price.
(b) What is the average variable cost (AVC) of producing 250 thumbs?
(c) What is the variable cost (VC) of producing 100 thumbs?
(d) Suppose your fixed costs are $1000. What is your profit if you make and sell 550
thumbs for $4.10 each?
Linear Analysis
67
Worksheet #10
Breaking Even
Key Question
Suppose the company Trabsimp produces and sells espresso machines. For each of the following
situations, determine the quantity when Trabsimp breaks even (T R = T C). The selling price, the
marginal cost (M C), and the fixed cost (F C) are given.
I. Assume the selling price is $40 per espresso machine, marginal cost is always $15 per
espresso machine, and fixed cost is $50,000.
II. Assume the marginal cost is doubled to $30 per espresso machine with the selling price and
fixed cost remaining the same as in question I.
III. Assume the selling price and the marginal cost are the same as in question I. However, the
fixed cost is reduced to $35,000.
IV. Assume the selling price is $24 per espresso machine, the marginal cost is a constant $10
per espresso machine, and fixed cost is $42,000.
1
How can you tell by reading Key Question I that the graphs of T R and T C will be straight
lines?
Key Question I
240
220
200
This graph shows T R and T C for Key Question I, where q is measured in thousands of
machines and T R and T C are measured in
thousands of dollars.
thousands of dollars
180
160
140
120
100
80
60
40
20
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Quantity (thousands of machines)
69
70
2
W ORKSHEET #10
Breaking Even
(a) In the graph for Key Question I, identify which line represents T R and which represents
T C. Explain your choice.
(b) Use the graphs to find the quantity at which Trabsimp breaks even in Key Question I.
(NOTE: It is unfortunate that the phrase break even is used in two different ways in
business mathematics. In this course, the breakeven price is the lowest market price at
which a business is not forced to take a loss, as described in previous worksheets. On
the other hand, a company will break even at a quantity q if total revenue is equal to
total cost at that quantity; i.e., if the T R and T C graphs intersect at that quantity. )
3
Draw the graph of T C for Key Question II on the same set of axes as Key Question I. Use
the graphs to find the quantity at which Trabsimp breaks even in Key Question II. Since the
value of M C was doubled in Key Question II as compared to Key Question I, some people
may conclude that the quantity at which the company breaks even would double as well. Is
that the case?
4
Draw the graph of T C for Key Question III on the following axes. (The graph of T R is
already given.) Use the graphs to find the quantity at which Trabsimp breaks even in Key
Question III.
Key Question III
220
200
thousands of dollars
180
160
140
120
100
80
60
40
20
0
.
T R....
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220
200
180
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100
80
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0
6
Quantity (thousands of machines)
5
240
thousands of dollars
240
Key Question IV
0
1
2
3
4
5
6
Quantity (thousands of machines)
Draw the graphs of T R and T C vs. quantity for Key Question IV on the indicated axes. Use
your graphs to find the quantity at which Trabsimp breaks even.
W ORKSHEET #10
Breaking Even
71
We’ll now answer the same Key Questions using functional notation and formulas. We’ll find a formula
for total revenue and total cost in each situation in the Key Questions. For example, T RI (q) will represent
the total revenue for selling q thousand machines in Key Question I.
Since the selling price and marginal cost are both constant, T R and T C are linear functions in each Key
Question. A linear function f (q) has the form f (q) = mq + b, where m is the slope and b is the vertical
intercept. The selling price is the slope of the T R graph. (Why?) Further, if q = 0, then T R = 0. (Why?)
So, in each Key Question, the vertical intercept of the T R graph is 0; T R will go through the origin. The
value of M C is the slope of the T C graph. (Why?) Further, if q = 0, then T C is equal to the value of fixed
cost.
So, in Key Question I, we have
T RI (q) = 40q and T CI (q) = 15q + 50.
For example, if we sell 3 thousand machines, then we plug 3 in for q in each formula:
T RI (3) = 120 thousand dollars and T CI (3) = 95 thousand dollars.
To find the quantity at which we break even, we set total revenue equal to total cost and solve for q:
40q = 15q + 50
25q = 50
q = 2.
This is the same answer that we found graphically in exercise #2
6
(a) Find the formulas for total revenue and total cost for Key Question II: T RII (q) and
T CII (q).
(b) Use algebra to find the quantity at which Trabsimp breaks even. Compare this to your
answers to exercise #3.
7
Repeat exercise #6 for Key Question III and Key Question IV.
8
Consider the following new situation:
The selling price is $p per machine, marginal cost is always $m per machine, and
fixed cost is c thousand dollars.
Repeat exercise #6 with this new situation.
9
If you did the last exercise correctly, then you found that the quantity at which Trabsimp
breaks even is given by
c
,
q=
p−m
where c =fixed cost, p =selling price, and m =marginal cost. (Why can we always assume
that p will be bigger than m so that this will never give us a negative number?)
72
W ORKSHEET #10
Breaking Even
(a) This is a general formula for the quantity at which Trabsimp breaks even. Check that
this gives the same answers as before for the Key Questions. (For example: In Key
50
= 2. Yes!)
Question I, c = 50, p = 40, and m = 15. So, q = 40−15
(b) Indicate whether this value of q gets larger or smaller under each of the following
circumstances.
i.
ii.
iii.
iv.
v.
vi.
selling price p goes down;
marginal cost m goes down;
fixed cost c goes down;
selling price p goes up;
marginal cost m goes up;
fixed cost c goes up.
10 Suppose you are in the situation described in Key Question I. The cost to produce each
machine is suddenly doubled so that the value of marginal cost is now $30 per machine.
How should you change your selling price so that the quantity at which you break even
doesn’t change?
While solving problems about breaking even, we have used some important algebraic skills. We have
• written down the formula of a linear function from a verbal description;
• graphed a linear function from its formula;
• solved a single linear equation in one unknown.
Worksheet #11
A Car with a Formula
Key Question
The graph below gives distance (D) vs. time (t) for a car. Also we know that the distance D at
time t is given by the formula
D(t) = t − 0.025t2 .
Use the graph and then use algebra to find each of the following:
I. the time t such that the average trip speed of the car at time t is 0.7 mpm;
Distance (miles)
II. a time t such that the average speed over the 5-minute interval starting at t is 0.4 mpm.
1
10
9
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10
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14
16
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20
Time (mins)
Use the graph to find each of the following:
(a) the distance the car travels in the first 16 minutes;
(b) the height of the graph at t = 9;
(c) the average trip speed after 14 minutes;
D(10)
(d)
;
10
(e) the slope of the secant line from t = 6 to t = 11;
D(20) − D(15)
(f)
5
73
74
2
W ORKSHEET #11
A Car with a Formula
Use the graph to find D(11), the car’s distance after 11 minutes. Then, use the formula
D(t) = t − 0.025t2 to compute D(11):
D(11) = 11 − 0.025 · 112 =
.
Do you get the same value from both methods?
3
Complete the last two rows of the following table. Compare the Numerical Values with your
answers to exercise #1 (a) and (b).
English
Graphs
Functional Notation
Formula or Numerical Value
the distance covered
by time t
the height of graph at
time t
D(t)
t − 0.025t2
the distance covered
by time 5
the height of the graph
at t = 5
D(5)
5 − 0.025 · 52 = 4.375 miles
the distance covered
by time 16
the height of the graph
at t = 9
Recall that the car’s average trip speed at time t is
distance traveled since t = 0
.
time elapsed since t = 0
Since the car travels D(t) miles in t minutes, average trip speed at time t is
D(t)
mpm.
t
The car’s average trip speed is read from the graph as the slope of a diagonal line.
W ORKSHEET #11
4
A Car with a Formula
75
Complete the last two rows of the following table. Compare the Numerical Values with your
answers to exercise #1 (c) and (d).
English
Graphs
Functional Notation
ave trip speed at time
t minutes
slope of diagonal line
to distance graph at t
D(t)
t
ave trip speed at time 6
minutes
slope of diagonal line
to distance graph at
t=6
D(6)
6
Formula or Numerical Value
t − 0.025t2
t
(6) − 0.025 · 62
= 0.85 mpm
6
ave trip speed at time
14 minutes
D(10)
10
5
Use the graph to answer Key Question I.
To answer Key Question I using algebra, it will help if we simplify the formula for average trip speed.
ave trip speed D(t)
t − 0.025 · t2
t 0.025t2
at time t =
=
= −
= 1 − 0.025t.
t
t
t
t
(Do you follow all the steps here?)
6
Use this new formula to compute the average trip speed at t = 10 and at t = 14. Compare
your answers to the table in exercise #4.
Answering Key Question I using algebra is now in our grasp. We need to find when average trip speed is
0.7 and we know average trip speed is 1 − 0.025t. So, we need to solve the equation
1 − 0.025t = 0.7.
7
Solve the equation 1 − 0.025t = 0.7 for t to answer Key Question I. Compare with your
answer to exercise #5.
Recall how to compute the car’s average speed during the interval from t = a to t = b. Average speed is
given by
distance traveled
.
time elapsed
Since the car travels D(b) − D(a) miles in b − a minutes, the car’s average speed is
D(b) − D(a)
mpm.
b−a
The car’s average speed is read from the distance graph as the slope of a secant line from t = a to t = b.
76
8
W ORKSHEET #11
A Car with a Formula
Fill in the last two rows of the following table. Leave the right-most column (Formula or
Numerical Value) blank for now.
English
Graphs
Functional Notation
average speed from
time t to time t + 5
minutes
slope of secant line
from t to t + 5
D(t + 5) − D(t)
5
average speed from
t = 4 to time t = 9
minutes
slope of secant line
from t = 4 to t = 9
D(9) − D(4)
5
Formula or Numerical Value
slope of the secant
line from t = 6 to
t = 11
D(20) − D(15)
5
9
Use the distance graph given at the beginning of the worksheet to answer Key Question II.
To complete the table from exercise #8 and answer Key Question II with algebra, we’ll need a formula for
D(t + 5) − D(t)
.
5
We already know that D(t) is t − 0.025t2 . We also need to know what D(t + 5) is. Look over exercise #2
again. To compute D(11) using the formula, you let 11 take the place of every t in D(t) = t − 0.025t2 . To
compute D(16) (in exercise 3), you let 16 take the place of every t in D(t) = t − 0.025t2 . To compute D(9),
you let 9 take the place of every t in D(t) = t − 0.025t2 . So, to compute D(t + 5), you should let “t + 5”
take the place of every t in D(t) = t − 0.025t2 :
D(t + 5) = (t + 5) − 0.025 · (t + 5)2 .
W ORKSHEET #11
A Car with a Formula
77
We can simplify this expression. (Make sure you understand how to do all of the following steps. It will be
your turn soon.)
(t + 5)2 = (t + 5)(t + 5) = t2 + 5t + 5t + 25 = t2 + 10t + 25
0.025 · (t + 5)2 = 0.025 · (t2 + 10t + 25) = 0.025t2 + 0.25t + 0.625
(t + 5) − 0.025 · (t + 5)2 = (t + 5) − (0.025t2 + 0.25t + 0.625)
= t + 5 − 0.025t2 − 0.25t − 0.625
= −0.025t2 + 0.75t + 4.375
We now have a simplified formula for D(t + 5):
D(t + 5) = −0.025t2 + 0.75t + 4.375.
10 Compute D(7) in two ways.
(a) Plug t = 7 into the formula for D(t).
(b) Plug t = 2 into the simplified formula for D(t + 5).
Even though we’ve already done a lot of work, we’re still not ready to answer Key Question II. We’re still
trying to find a simplified formula for
D(t + 5) − D(t)
.
5
So far, we have D(t + 5) = −0.025t2 + 0.75t + 4.375 and D(t) = t − 0.025t2 .
11
(a) Find a formula for D(t + 5) − D(t). Simplify as much as you can.
D(t + 5) − D(t)
. Simplify as much as you can. (It may help you
5
to remember that dividing by 5 is the same as multiplying by 51 .)
(b) Find a formula for
D(t + 5) − D(t)
, go back and fill in the last column
12 Now that you have a nice formula for
5
in the table in exercise #8. For the last two rows, you can compare your Numerical Values
to your answers to exercise #1 (e) and (f).
13 Use the formula you found in exercise #11 (b) to answer Key Question II. (Note: Some
people have trouble deciding what to do with the 0.4 in the Key Question: should they plug
0.4 in for t or set the formula equal to 0.4 and solve for t? If you are having trouble, read
the Key Question very carefully. Is 0.4 a time or a speed? Are you looking for a time or a
speed? The answers to those questions should help you make the correct choice.)
78
W ORKSHEET #11
A Car with a Formula
Notice that the formula that we started with, D(t) = t − 0.025t2 , is quadratic. (That is, D(t) is of the
form at2 + bt + c, where a = −0.025, b = 1, and c = 0.) We found a formula for the car’s average trip
speed
D(t)
= 1 − 0.025t
t
and a formula for the car’s average speed over a five-minute interval starting at time t
D(t + 5) − D(t)
= 0.875 − 0.05t.
5
The two speed formulas are both linear. (That is, each speed is of the form mt + b. For the average trip
speed, m = −0.025 and b = 1. For the average speed over a 5-minute interval, m = −0.05 and b = 0.875.)
14 Suppose the distance formula for a different car is D(t) = 2t − 0.04t2 .
(a) Find a linear formula for the car’s average trip speed at time t.
(b) Find the time at which the car’s average trip speed is 1.32 mpm.
(c) Find a linear formula for the car’s average speed from time t to time t + 2.
(d) Find the time t such that the car’s average speed over the 2-minute interval starting at t
is 1.04 mpm.
→
15 The parabola below has a formula that looks like f (x) = ax2 + bx + 8.
a) Use the fact that f (2) = 8, and f (4) = 0 to
figure out the actual values of a and b.
b) Evaluate
f (3)
f (3) − f (0)
and
.
3
3
8
c) Write out a formula for
f (x) − f (0)
.
x
6
4
d) Evaluate f (3.5) − f (3).
e) Write out a formula for f (x + 2) in the
form: f (x + 2) = ( )x2 + ( )x + ( ).
f) Write
out
the
f (x + 2) − f (x)
.
2
formula
for
2
.
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W ORKSHEET #11
→
A Car with a Formula
16 The graph to the right gives the amount
of water that has flowed into a reservoir
by time t. The formula for this graph is
w = I(t) = at2 + bt (where w is in gallons
and t is in hours).
a) Write out a formula in terms of a, b,
and t for the overall average rate of
).
flow into the reservoir (i.e., I(t)
t
= 15 and I(10)
= 10,
b) Given that I(5)
5
10
determine the values of a and b.
79
w
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t
c) Find the amount of water that flows into the reservoir from time t = 3 hours to time
t = 4 hours.
d) Water flows out of the reservoir at a constant rate of 8 gallons per hour. The reservoir
has 10 gallons in it at time t = 0. Write out a formula for the actual amount of water
contained in the reservoir at time t.
→
17 Two balloons are released at the same time.
One goes up steadily, and the other goes
up for a while and then begins to descend.
The formulas for the heights h (feet) of the
balloons at time t (minutes) are:
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Balloon A : h = A(t) = − 2 t + 15t
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Balloon B : h = B(t) = 5t.
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By the Average Rate of Ascent at time t, .... ... ......
.... ...
I mean the overall average vertical speed ..........................................................................................................................................
from time 0 to time t (i.e., A(t)
or B(t)
).
t
t
a) Write out the formulas for Average Rate of Ascent of the two balloons at time t.
b) At what time will the two balloons have the same Average Rate of Ascent?
c) Find the time at which the Average Rate of Ascent of A exceeds the Average Rate of
Ascent of B by 7 ft/min.
d) Find the Average Rate of Ascent of A when B is 40 ft high.
Worksheet #12
The Changing Price List at the Print Shop
Key Question
Once again, suppose you are running the print shop as in Worksheet #3. With the price list given
in that worksheet and a cost of paper of $4 per ream, we found (after much labor) that you make a
profit for all orders in the range 0 to 12 reams. Profit was maximized at q = 6 reams of paper.
(I) If the cost of paper goes up by $2 a ream and the price list remains the same, then what is
the quantity that maximizes profit?
(II) If you raise your prices by $1 “across-the-board” and the cost to you remains $4 a ream, then
what is the quantity that maximizes profit?
Before we get to the algebra, let’s review the process that we used to maximize profit in Worksheet #3.
The process went as follows:
• Based on the initial verbal description of the price list, you drew the straight-line graph of price vs.
quantity. (See the graph labeled P L below.)
• Using the fact that total revenue = p × q, you compiled a table of the values of total revenue. Then,
you drew the T R-graph. (See the graph labeled T R below.)
• Using the cost of paper ($4 a ream), you calculated the values of total cost and drew the T C-graph.
(See the graph labeled T C below.)
• Using a clear plastic ruler to measure the vertical distance between the T R and T C graphs, you
found the quantity (q = 6) where profit is greatest.
• You checked the value q = 6 using “marginal analysis”: For quantities up to 5 reams, marginal
revenue is greater than the cost of an additional ream of paper. So, increasing quantity by 1 ream
increases your profit. But, starting at 6 reams marginal revenue is less than the cost of an additional
ream of paper. So, increasing quantity by 1 ream decreases profit. This means that profit is maximized
at q = 6 reams.
80
W ORKSHEET #12
The Changing Price List at the Print Shop
80
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2 4 6 8 10 12 14 16
70
60
50
40
30
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10
81
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............................................................................................................................................................................................................. q
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
We can now use algebra and marginal analysis to maximize profit.
1
The graph of price vs. quantity (P L) is a line. Find the equation of this line.
2
Recall that we compute total revenue by multiplying price by quantity. Use the equation you
found in the previous exercise to find a formula for T R(q), total revenue at q reams.
3
Use the formula to compute total revenue at q = 3 reams and at q = 12 reams. Check your
answers against the table in Worksheet #3, exercise #2, and against the graph of T R given
above.
We’d like to develop a formula for marginal revenue. Recall that marginal revenue at q is the change in
total revenue as we increase quantity from q to q + 1. That is,
M R(q) = T R(q + 1) − T R(q).
We’ll use the formula for T R from exercise #2 to find a formula for M R(q).
The formula for T R is
T R(q) = 16q − q 2 .
If you did exercise #3 correctly, then, in order to compute T R(3), you replaced each q in the T R formula
with a 3:
T R(3) = 16(3) − 32 = 39.
82
W ORKSHEET #12
The Changing Price List at the Print Shop
To compute T R(12), you replaced each q with a 12:
T R(12) = 16(12) − 122 = 48.
So, in order to compute T R(q + 1), we need to replace each q with q + 1:
T R(q + 1) = 16(q + 1) − (q + 1)2 .
We expand and simplify (like in Worksheet #11):
T R(q + 1) = 16(q + 1) − (q 2 + 2q + 1)
= 16q + 16 − q 2 − 2q − 1
= −q 2 + 14q + 15.
This gives us T R(q + 1), but what we’re looking for is M R(q):
M R(q) = T R(q + 1) − T R(q)
= (−q 2 + 14q + 15) − (16q − q 2 )
= −q 2 + 14q + 15 − 16q + q 2
= 15 − 2q.
4
Use this formula to compute marginal revenue at q = 2 and at q = 8. Check your answers
against the table in Worksheet #3, exercise #12, and against the graph of M R given at the
beginning of this worksheet.
5
Recall that profit increases as long as M R is larger than our cost per ream of $4; profit
decreases once M R drops below $4. Use the formula for M R to find the value of q at which
M R is equal to 4. Check your answer against the graph of M R at the beginning of this
worksheet.
Since M R is a line with negative slope (m = −2), the values of M R decrease as quantity goes up. If
M R = 4 at q = 5.5, then: for quantities smaller than 5.5, M R is bigger than 4; for quantities larger than
5.5, M R is smaller than 4. (Do you see this in the graph of M R?) This means profit increases as quantity
changes from 0 to 1, from 1 to 2, and so forth, on up to from 5 to 6 (since M R at q = 5 is still bigger than
$4). But, starting at q = 6, M R is less than 4, so profit decreases as quantity changes from 6 to 7, from 7
to 8, etc. Do you understand why this means that profit is maximized at q = 6, the first whole number after
5.5?
6
Answer Key Question I. (HINT: If the price list stays the same, then your formulas for total
revenue and marginal revenue do not change.)
Now let’s work on Key Question II.
7
You’re raising prices by $1 per ream at every quantity. So, an order of 1 ream will be priced
at $16 per ream, 2 reams will be priced at $15 per ream, 3 reams will be priced at $14 per
ream, etc. Sketch the graph of price vs. quantity on the axes on the left below and find the
equation of the line you drew.
W ORKSHEET #12
The Changing Price List at the Print Shop
83
8
Find a formula for total revenue, T R(q), based on this new price list. Sketch the graph of
T R on the axes on the right below. (The total cost graph is the same as before.)
9
Find a formula for marginal revenue, M R(q). (Mimic the procedure we executed in finding
the M R formula for the original price list.) Sketch the graph of M R on the axes on the left
below.
80
16
14
12
10
8
6
4
2
p
.....
..
..
..
..
...
..
..
..
..
..
..
..
..
..
................................................................................................
q
2 4 6 8 10 12 14 16
70
60
50
40
30
20
10
.$
......
..
..
...
..
...
..
...
..
...
..
.
.
.
.
..
..
.... T C
.
.
.
..
.
....
..
.
.
.
.
..
....
..
.
.
.
.
..
....
..
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.. ......
............................................................................................................................................................................................................. q
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
10 Find the quantity that maximizes profit. You can do this three ways:
(a) Use the graphs of T R and T C to find the quantity at which the vertical distance between
T R and T C is greatest.
(b) Use the graph of M R to find the quantity at which the graph of M R has height 4. If
this is not a whole number, then select the next whole number of reams.
(c) Use algebra to find where M R = 4. If this is not a whole number, select the next whole
number of reams.
You just answered Key Question II.
Let’s investigate the last two methods of the previous exercise a bit further. Recall that marginal cost at q
is defined as the change in T C as quantity changes from q to q + 1. Since the cost of paper is $4 per ream,
increasing an order by 1 ream always increases our T C by $4. That is, M C(q) = 4 no matter what q is.
11 Compute marginal cost at q = 1, q = 2, and q = 3. Sketch the graph of M C on the axes
with M R on the left above.
84
W ORKSHEET #12
The Changing Price List at the Print Shop
We can now restate methods (b) and (c) in exercise #10 as:
b) Use the graphs of M R and M C to find the quantity at which the graph of M R intersects the graph
of M C. If this is not a whole number, select the next whole number of reams.
c) Use algebra to find where M R = M C. If this is not a whole number, then select the next whole
number of reams.
In other words, under certain conditions, we maximize profit when marginal revenue equals marginal cost.
12 We end this worksheet by giving you one more opportunity to practice what you’ve learned.
Suppose the cost of paper drops to $3.25 per ream. You alter your price list as follows:
$11.50 for an order of 1 ream,
$11.00 for an order of 2 reams,
$10.50 for an order of 3 reams,
$10.00 for an order of 4 reams,
and so forth. Find formulas for price p, total revenue T R, marginal revenue M R, and
marginal cost M C. Use algebra to find the quantity that maximizes profit.
Summary:
• If you have a formula for price p (in terms of quantity q), then the formula for total revenue os given
by
T R(q) = p × q.
• The formula for marginal revenue is given by
M R(q) = T R(q + 1) − T R(q),
which can usually be simplified.
• Profit is maximized at the first quantity at which M R falls below M C (in the case of the print shop,
M C was simply cost per ream). If this quantity is not a whole number, we go up to the next whole
number.
→
13 The graphs to the right are of total cost
and total revenue of manufacturing Trivets.
Their formulas are:
TR : R(q) = − 14 q 2 + 30q
TC : C(q) = 17.5q + 100.
.
......
.............
.
.
.
.
.
.
..
.
.
...
..
TR ...................
.
.
..
.
..
............... TC
.
.
.
.
..
.. ............
.......................................................................................................
.
q
Recall that
M R(q) = T R(q + 1) − T R(q) and M C(q) = T C(q + 1) − T C(q).
W ORKSHEET #12
The Changing Price List at the Print Shop
85
a) Find formulas for M R and M C.
b) Find the quantity q at which profits are maximized.
c) Find the quantity q at which MR exceeds MC by $5.25.
d) Find the quantity q at which Average Revenue is $16 per Trivet.
e) Find the quantity q at which Average Cost is $21.50 per Trivet.
→
14 The total revenue from manufacturing and selling q Blivets is given by the formula:
√
TR(q) = 5 q
where q is given in thousands of Blivets, and T R(q) is given in thousands of dollars.
a) Suppose m is a particular quantity (in thousands). Write out a formula in terms of m
for the Marginal Revenue of producing m thousand Blivets. Do not simplify.
b) The total cost of manufacturing q thousand Blivets is given by the formula: T C(q) =
q+1 (thousand dollars). Write out the equation you would solve to answer the following
question.
What is the largest quantity of Blivets you can manufacture and not take a
loss?
Do not solve the equation.
c) Find the value of q at which the average cost of manufacturing Blivets is $1.25 per
Blivet.
d) Find the value of q at which the average revenue for manufacturing Blivets is $1.30 per
Blivet.
e) Write out the equation that you would solve to answer the following question:
For what quantities q is the profit of manufacturing q thousand Blivets going
to be $4,000?
Do not solve the equation.
86
W ORKSHEET #12
The Changing Price List at the Print Shop
Quadratic Analysis
87
Worksheet #13
Two Vats of Water
Water is flowing in and out of two vats, a red vat
and a green vat. The graphs to the right show the
amounts of water in the two vats over an eighthour period. We also have a gauge which measures the difference:
Amount in Green Vat
minus
Amount in Red Vat.
Note that the gauge starts out negative, is then
positive for several hours, and ends up negative.
Gallons
Introduction
28
24
20
16
12
8
4
..... w
..
......... .........................
...
......
.
.. .........
.
R....
.....
.. .......
.. .. .....
.
.
.
.... ..... ..
.......
.....
. ..
............
..
..............................
..
...
..
...
..
..
...G
..
..
..
...............................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 t
Hours
Key Questions
I. Find the time at which the water in the red vat hits its lowest level.
II. Find the actual lowest value the water level in the red vat reaches.
III. Find the time at which the difference gauge reaches its highest value.
IV. Find the actual highest value the difference gauge reaches.
V. Find the first time at which the water levels in the two vats are the same.
1
Use the graph to estimate each of the quantities in the Key Questions. We’ll use algebra to
find their exact values.
The formula for the amount of water in the red vat is
1
R(t) = t2 − 5t + 25;
2
the formula for the amount of water in the green vat is
G(t) = −t2 + 6t + 16.
Both of these are quadratic functions. (You may want to review the section on quadratic functions in the
Prologue.)
89
90
W ORKSHEET #13
Two Vats of Water
2
Key Question I requires finding the t-value of the vertex of the graph of R(t). Use the vertex
formula to find the time at which the water level in the red vat is lowest.
3
Key Question II requires finding the R-value of the vertex of the graph of R(t). You just
found the t-value that gives the lowest level. Plug this value in for t in the formula for R(t)
to find the lowest level in the red vat.
4
To answer Key Question III, we need a formula for the difference, D(t). At the start of this
worksheet, we said that our difference gauge measures the amount in the green vat minus
the amount in the red vat. So,
D(t) = G(t) − R(t).
We have formulas for G(t) and R(t). So,
1
D(t) = (−t2 + 6t + 16) − ( t2 − 5t + 25).
2
Collect like terms and get a simple formula for the difference:
D(t) = ( )t2 + ( )t + ( ).
The difference formula is another quadratic. Does its vertex give a high point or a low point?
How can you tell? Use the vertex formula to answer Key Question III.
5
Answer Key Question IV.
6
Key Question V requires that we find when R(t) and G(t) are the same. That is, we need to
know the values of t at which R(t) = G(t). We have formulas for R(t) and G(t). So, we
need to solve the equation
1 2
t − 5t + 25 = −t2 + 6t + 16
2
for t. This is a quadratic equation. To solve it, write it in the form at2 + bt + c = 0 and
use the quadratic formula. If you do this correctly, you will get two different values of t.
Re-read Key Question V to choose the desired time.
7
Check your answers to exercises #2 to 6 against the estimates you made in exercise #1.
Now imagine that we can control the flow of water in and out of the red vat in two ways:
Amount Shift: We keep basically the same red vat graph, but start with either more or less water, so that the
graph is shifted up or down.
Time Shift: We keep basically the same red vat graph, but make everything happen earlier or later, so that
the graph is shifted to the left or to the right.
W ORKSHEET #13
8
Two Vats of Water
91
Fill in the “No Shift” column of the following table with your answers from exercises #2 to
6. Then, in the “Change?” column for each type of shift, enter either a yes or a no, indicating
whether you think your answer to each Key Question will change after we shift the red graph
in the indicated direction. Leave the remaining two columns blank for now.
No Shift
KEY QUESTION
Answer
Amount Shift
Time Shift
(R shifts up or down)
(R shifts left or right)
Change?
New Answer
Change?
New Answer
I. Find the time at which the water in the red vat hits its lowest
level.
II.
Find the actual lowest value
the water level in the red vat
reaches.
III.
Find the time at which the
difference gauge reaches its
highest value.
IV. Find the actual highest value
the difference gauge reaches.
V. Find the first time at which the
water levels in the two vats are
the same.
9
Suppose we started with 5 more gallons in the red vat. This would give us a new function
which we could denote by R5 (t). (The subscript “5” tells how much additional water there
is.) This means that there are always 5 gallons more in the red vat than we had in the “No
Shift” situation. Thus R5 (t) = R(t) + 5. Use this to get a formula for R5 (t). What is the
formula for R15 (t), the amount of water in the red vat if we start with 15 gallons more? If
we start with 10 fewer gallons, then it makes sense to call the resulting function R−10 (t).
What is the formula for R−10 (t)? What is the relation between the graphs of R(t), R5 (t),
and R−10 (t)?
Are you ready to agree with the following assertion?
If the red vat starts with d more gallons in it (where negative d
means “d fewer”), then at time t the red vat has
1
Rd (t) = t2 − 5t + 25 + d gallons in it.
2
10 Write a formula (which will involve d) for Dd (t), the reading on the difference gauge at time
t, if the red vat starts with an additional d gallons in it. (Note that we’re not changing what
happens to the green vat, so the formula for G(t) doesn’t change.)
92
W ORKSHEET #13
Two Vats of Water
Now that we have the formulas for Rd (t) and Dd (t), we should be able to use our knowledge of the quadratic
and vertex formulas to tell whether an Amount Shift (up or down) changes the answers to the five Key
Questions. But, in fact, we can do better than that—we can actually answer these questions and see if
throwing the d into the formula makes any difference.
11 At what value of t does the function Rd (t) = 12 t2 − 5t + 25 + d reach its vertex? What is
the height of the vertex? Enter these answers in the appropriate places in the table of Key
Questions and check whether an Amount Shift has made a difference or not.
12 Using the formula for Dd (t), find the specific answers to Key Questions III and IV under an
Amount Shift of d gallons. Enter these answers in the table of Key Questions.
13 To answer Key Question V under an Amount Shift of d gallons, we need to solve the equation
1 2
t − 5t + 25 + d = −t2 + 6t + 16.
2
If we collect all the terms on one side of the equation, we get
3 2
t − 11t + (9 + d) = 0.
2
Use the quadratic formula with
3
a = , b = −11, and c = 9 + d
2
to answer Key Question V under an Amount Shift of d gallons and check whether the shift
changes your answer from the “No Shift” case.
So, for an Amount Shift, the algebra is not bad and it confirms easily what the graphs told us.
We need a formula for the amount of water in the red vat after a Time Shift. As with the Amount Shift,
we would like a general formula that results from a shift of h hours earlier or later. I will lead you to such
a formula, but if you begin to get lost, try the particular case of h = 1 (everything is one hour later) or
h = −2 (everything is 2 hours earlier). To keep things straight, we will denote by Rh (t), the amount of
water in the red vat at time t, if everything happens h hours later. The superscript h tells us how many hours
later everything happens.
14 On the axes at the beginning of the worksheet, draw a rough sketch of the graph of R1 (t)—
the graph of the red vat’s water level if everything happens in the red vat one hour later. (This
is exactly like the Lagging Car of Worksheet #5.)
15 In trying to write a formula for R1 (t) it is tempting to look at the formula for R(t) and “add”
or “attach” the number 1 somewhere. The most obvious place to do this is to add the 1 on
the end of the formula:
1
R(t) + 1 = t2 − 5t + 25 + 1.
2
If you do this, you won’t be writing the formula for a time shift of 1 hour. What does this
formula represent instead?
W ORKSHEET #13
Two Vats of Water
93
To find the formula we are searching for, we have to use functional notation first. We want Rh (t)—that is,
the amount of water in the red vat if everything happens h hours later. This expression, “everything happens
h hours later” means to me “The amount of water in the red vat at time t under the new condition is the
same as the amount of water in the red vat under the original (“No Shift”) condition at time t − h.” This
means
Rh (t) = R(t − h).
16 In exercise #14, you drew the graph of R1 (t). On the graph, verify that R1 (1) = R(0),
R1 (2) = R(1), R1 (5) = R(4), and R1 (10) = R(9). (There’s nothing to show here. Just
make sure you understand what we’re asking you to do and do it.)
17 Replace each t in the R(t) formula with a t − h to get a formula for Rh (t):
Rh (t) = R(t − h) =
.
Don’t simplify.
Let’s simplify this formula together. You should have that
1
Rh (t) = (t − h)2 − 5(t − h) + 25.
2
We can expand (t − h)2 as:
so that
Next, we distribute the
(t − h)2 = t2 − 2ht + h2 ,
1
Rh (t) = (t2 − 2ht + h2 ) − 5(t − h) + 25.
2
1
2
and the −5 to get
1
1
Rh (t) = t2 − ht + h2 − 5t + 5h + 25.
2
2
Finally, we collect like terms and write Rh (t) in the form at2 + bt + c:
1
1
Rh (t) = t2 − (5 + h)t + ( h2 + 5h + 25).
2
2
18 The formula for Rh (t) is a quadratic function of the variable t. But it looks strange because
the coefficients have h’s in them. If Rh (t) is a quadratic function in the form Rh (t) =
at2 + bt + c, then what are a, b, and c? (I’ll give you the easy one: a = 12 .)
19 Use the vertex formula to answer Key Question I under a Time Shift. Does the Time Shift
give a different answer than the “No Shift” case? Does this new answer make sense when
you consider that the Time Shift shifts the red graph to the left or right by h hours?
94
W ORKSHEET #13
Two Vats of Water
20 To answer Key Question II with algebra, you would plug your answer to Key Question I into
the formula for Rh (t). Yuck! Instead, use the graphs to explain why the Time Shift does not
change the answer to Key Question II — it will be the same as in the “No Shift” case. (If
you really want to practice your algebra, go ahead and plug t = 5 + h into the formula for
Rh (t) and expand and simplify until you get 12.5.)
Again, if we tried to use algebra to answer Key Questions III, IV, and V, for a general Time Shift of h
hours, things would get ugly pretty quickly. Instead, we’ll answer these Key Questions with the specific Time
Shift of h = 1 hour.
21 Recall the formula for Rh (t):
1
1
Rh (t) = t2 − (5 + h)t + ( h2 + 5h + 25).
2
2
Replace each h with a 1 to obtain the formula for R1 (t). Use this formula and some algebra
to answer Key Questions III, IV, and V under a Time Shift of h = 1 hour. Check whether
the Time Shift changes your answers from the “No Shift” case.
→
.....
18 ..
........
.
..
.
22 The graph to the right is of distance vs. time
.
.
...
..
for a car, Car F. Its formula is F (t) = 3t −
.
..
.
..
1 2
..
t where t is in minutes and distance is in
..
.
8
.
.. ..
miles. We will call the place where Car F
.. ..
.. ..
is at 0 minutes the Starting Place.
.. ..
.. ..
....
a) How long does it take the car to go
.....................................................................
.
.
the first 5 miles?
12
b) How long does it take the car to travel from a point 2 miles from the Starting Place to
a point 7 miles from the Starting Place?
c) Another car, Car G, has a linear distance vs. time graph about which we have the
following information. Car G is 5 miles ahead of Car F at time t = 0, and Car F
passes Car G at 9 minutes. Give the formula for G(t), the distance from the Starting
Place for Car G at time t.
d) How far is Car G from the Starting Place at time t = 10?
e) How far apart are the two cars at time t = 4?
f) At what time in the first 9 minutes are the two cars 2 miles apart?
g) Suppose another car, Car H, travels at exactly the same speed as Car F, but is 1 mile
ahead of Car F at time 0.
(i) Give the formula for H(t), the distance that Car H is from the Starting Place at
time t.
W ORKSHEET #13
Two Vats of Water
95
(ii) How far apart are Car H and Car F at time t = 9.67 minutes?
→
23 The graph to the right is of the amount of
water in a vat at time t. It is given by the
formula
A(t) = t2 − 8t + 20,
where t is in minutes and A is in gallons.
a) At what time does the water reach its
lowest level? How many gallons are
in the vat at that time?
A
......
....
..
.. ..
.
.
.. ...
..
.. ...
.
.
.. ....
..
.
.
..
.
.
.
.
.
......... .......
..
........
..
.....................................................................................................
.
t
b) Give the two times when the vat has 10 gallons in it.
c) By how much is the amount of water in the vat changed from 6 to 8 minutes?
d) Another vat, Vat B, has a graph of amount vs. time which is a straight line. Water
comes into Vat B at a constant rate of 1/2 gallon per minute. If Vat B were empty at
time t = 0, would it ever have as much water in it as Vat A?
e) Suppose that Vat B actually starts with 8 gallons in it. At what time will the difference
B(t) − A(t) be greatest?
f) Name a time when the water level in Vat A is increasing and the difference B(t) − A(t)
is also increasing.
→
24 The graphs to the right are parabolas with
formulas:
f (x) = x2 − 4x + 16
g(x) = − 12 x2 + 4x + 10.
a) Find the value of g(3) − f (3).
b) Find the values of x at which the two
graphs cross.
.
.
.....
.
.
.
.
.
.
.
.
.
.
.
......... ..................... .................
.....
.. .......................
.....g(x)
.....
f (x)
..
....
..
....
..
...
..
.
..
.........................................................................................................................................
c) Find the values of x at which g(x) − f (x) = 4.
d) Find the longest interval you can, starting at x = 1, over which g(x) − f (x) is increasing. (Hint: First look for where g(x) − f (x) hits its peak.)
e) Find the longest interval over which both f (x) and g(x) are increasing. (Hint: Look
for where the two functions peak or bottom out.)
f) Find the set of all values for which g(x) is greater than or equal to 15.
96
→
W ORKSHEET #13
Two Vats of Water
25 Charlie throws a rock from a building. At time t (seconds) the height h (feet) above street
level is recorded. This height is given by the formula
h = F (t) = −16t2 + 30t + 20.
a) How high is the building Charlie is standing on?
b) At what time does the rock hit the ground?
c) At what time does the rock reach its greatest height?
d) Sketch the graph of h vs. t, indicating the rock’s high point and the time it hits the
ground.
e) At what time has the rock first traveled to a point 10 feet higher than the height at which
it was thrown?
f) At what time does the rock reach an elevation of 20 feet above street level and is on its
way down?
→
26 The formula for the average trip speed of a car is s(t) =
3
t
+ 6 feet/sec.
a) The recipe for distance-covered given average-trip-speed is
Distance covered = D(t) = t × s(t).
Write out a formula for D(t).
b) Find the distance covered from t = 3 to t = 7 seconds.
c) Find the value of t for which D(t) is 15 ft.
d) Another car has the following formula for average trip speed vs. time:
q(t) = 2t + 1.
(i) Write down the equation you would solve in order to find the time t when the two
cars have the same average trip speed.
(ii) Rewrite your equation from (i) in a form that can be solved using the quadratic
formula.
→
e) At what times are the two cars 5 feet apart? [Note: there are three answers!]
.
.....
27 To the right is the graph of the quadratic
..
...........................
.. ............
function p(x) = −x2 + 4x + 12. Let q(x)
.......
.. ....
......p(x)
be a different function defined by the recipe
......
....
..
p(x)
....
..
.
q(x) =
..
....
x
..
..
..
a) Find the value of q(3).
...
..
..
b) Find a formula for q(x).
..
..
......................................................................................................................
W ORKSHEET #13
Two Vats of Water
97
c) Which of the graphs below looks like it could be the graph of q(x) from x = 1 to
x = 4? Justify your answer.
NONE
d) Find the positive value of x at which the value of q(x) is 5.
e) What is the largest value of q(x) for values of x in the interval x = 1 to x = 5? Show
your work and tell why your answer is correct.
f) (i) Set up the equation you would solve in order to answer the following question:
For what positive value of x does the graph of q(x) intersect the straight line graph
y = 2x + 4?
(ii) Rewrite the equation you gave in (i) into a form that can be solved by using the
quadratic formula.
(iii) Answer the question given in (i).
→
28 The two graphs to the right are of the
amounts of water in two vats, Vat A and
Vat B. The formulas for these graphs are as
follows:
√
Vat A : A(t) = 4t − t + 1
√
Vat B : B(t) = 12 t
(t is measured in hours, and amounts in
gallons.)
.
......
..............
.
.
.
..
.
.
.
.
.
..
.................
.
.
.
..
.
.
.
..
...... ........A(t)
.
.
..
.
.
.
.
.. B(t).....
....
.
.. .....
.
.
.
.. ..
....
.
.. ..
.
.
.
.... ....
.. ...
...............................................................................................................
t
a) What is the change in the amount of water in Vat A from time t = 2 to time t = 6?
b) The overall rate of change of water in Vat B is given by the recipe
overall rate of change of water in Vat B at t = 6 hours?
B(t)
.
t
What is the
c) Find the two times at which the amounts of water in the two vats are equal.
d) Write an equation that could be solved directly by the quadratic formula to answer the
following question:
At what time(s) are there 9 more gallons in Vat B than Vat A?
e) How much water is in Vat A when there are 36 gallons in Vat B?
Worksheet #14
The Area-of-the-Rectangle Function
Introduction
In the picture to the right we have a new way of
producing functions. For each value of x, we
go up to the straight-line graph and get a value
of y. The vertical and horizontal lines we draw
by doing this, when taken with the x-axis and yaxis, give us the shaded rectangle. We measure
the area of this rectangle. As we vary x, we get
different shaped rectangles with different areas.
Thus, this area is a function whose value depends
on x.
8
6
4
2
y ..
..... ....
..............
.. ............
x
.. .. .........
..................................•...... f (x) = − + 8
2
............................................................ .........
.
.
.
..............................................................
.
.
.
.
.
..........
..............................................................
..........
...................................................................
...................... ....
.................area
.
.
.
..........is
............
.
.
.
.
.
.........................................................
...............................
.................S(4)
.....................
.....
..........................................................
6
.......................................................... g(x) = + 2
..........................................................
x
..........................................................
.............................................................................................................................................................
x
2
4
6
8
10 12
We can repeat this way of producing areas with the curvy graph, instead of the straight-line
graph.
Let S(x)
be the area of the rectangle defined from the straight–line graph.
Let C(x) be the area of the rectangle defined from the curvy graph
Suppose we take some readings of the two functions S(x) and C(x). From the straight-line
graph, the rectangles start out tall and narrow, become more square-like and then get short and
wide. Thus the values of S(x) reach a peak somewhere in the middle. On the other hand, values
of C(x) calculated from the curvy graph just keep getting bigger. (We’ll investigate these claims
later in this worksheet.)
Now, what if for each x, we subtract the area obtained from the curvy graph from the area
obtained from the straight-line graph. Let D(x) be the difference in these areas. Then we have
D(x) = S(x) − C(x). As an example, you can see D(4), the area of the shaded rectangle in the
sketch below.
y.
. ..
...... ...
8 ..............
.. ........
.. .. .........
x
......................................•.... f (x) = − + 8
6 .............................................. .......
2
Key Question: Does the area given
........................................................................... ...........
.
.
..............................................................
..........
by D(x) reach its peak before or after
.
.
.
4 ..................................................................•.......
.
.
.
.
S(x)
does?
.
........................... ............
..
....
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.... ..............
....
6
.
2 ...
.
+
2
.... g(x) =
..
x
...
.
..
..................................................................................................................................................................
x
2
4
6
8
10 12
98
W ORKSHEET #14
The Area-of-the-Rectangle Function
99
The straight-line graph is given again on the left below. Draw rectangles, as described in the
introduction, at x = 2, 4, 6, and 8. Compute the areas of these rectangles. You have just
computed S(2), S(4), S(6), and S(8).
1
8
6
4
2
y
.....
.............
.. ..........
..........
..
..........
..
..
..........
..
..........
..
..........
..
..........
..
..........
..
.......
..
..
..
...
.............................................................................................................................................................
x
2
4
6
8
10 12
8
6
4
2
y ..
..... ....
.. ..
.. ..
.. ..
.. ..
...
..
..
....
..
......
..
............
..
................................
..
.....................
..
..
..
..
...............................................................................................................................................................
x
2
4
6
8
10 12
The formula for the straight-line graph is given as
f (x) = −
x
+ 8.
2
When you computed S(4), you found the area of a rectangle whose base is 4 units long and whose height is
6 units long. The height is the value of f (4):
4
f (4) = − + 8 = 6.
2
So, S(4) = base · height = 4 · f (4) = 4 · 6 = 24.
2
Do the same with x = 2, 6, and 8 to check your answers to exercise 1.
S(2) = 2 · f (2) = 2 · ( ) =
S(6) = 6 · f (6) = 6 · ( ) =
S(8) = 8 · f (8) = 8 · ( ) =
3
The curvy graph is given again on the right below exercise #1. Draw rectangles at x = 2, 4,
6, and 8 and compute C(2), C(4), C(6), and C(8).
4
The formula for the curvy graph is g(x) = x6 + 2. Use the formula for g(x) to check your
answers to exercise 3.
C(2) = 2 · g(2) = 2 · ( ) =
C(4) = 4 · g(4) = 4 · ( ) =
C(6) = 6 · g(6) = 6 · ( ) =
C(8) = 8 · g(8) = 8 · ( ) =
100
W ORKSHEET #14
5
Use your computations from the previous exercises to sketch rough graphs of
S(x) and C(x) from x = 2 to x = 12
on the axes at right.
6
These rough sketches should allow you
to answer the key question. Even
though you don’t have the graph of
D(x) = S(x) − C(x), you can see its
values as the vertical distance between
S(x) and C(x). Which function hits its
peak first: S(x) or D(x)? (Why aren’t
we asking when C(x) hits its peak?)
The Area-of-the-Rectangle Function
..... A
..
..
...
..
...
..
..
..
...
..
...
..
...
..
...
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.............................................................................................................................................................
x
2
4
6
8
10 12
With this area-of-the-rectangle function construction we have pushed pure graph reading techniques
about as far as they will go. The Key Question, which is not a difficult one, was almost impossible to
answer right from the first graphs you were shown at the beginning of the worksheet. With some labor we
could roughly sketch the graphs of S(x) and C(x) and make a good attempt at answering the Key Question.
But suppose the Key Question had been more precise:
Revised Key Question. At what value of x does S(x) reach its maximum value? At what value of x does
D(x) reach its maximum value?
Clearly, algebra is the way to answer these questions. We need formulas for S(x) and C(x).
7
Go back to your answers to exercises #2 and #4, where you found the values of S(2), S(6),
and S(8) from values of f (2), f (6), and f (8), and you found values of C(2), C(4), C(6),
and C(8) from values of g(2), g(4), g(6), and g(8). Use the same idea to find formulas for
S(x) and C(x):
S(x) = x · f (x) = x · (
)=
C(x) = x · g(x) = x · (
)=
Simplify the formulas if you can.
8
Find a formula for D(x). Again, simplify if you can.
9
S(x) and D(x) are both quadratic functions. So, their graphs are parabolas. For each, decide
whether the parabola opens up or down. Finally, answer the Revised Key Question.
W ORKSHEET #14
The Area-of-the-Rectangle Function
101
Perhaps some of these graphs and some of these computations have a familiar ring to them. Do you
remember when you have seen them before? In a slightly different language, from a slightly different point
of view, you have been doing the “Print Shop” problem here.
You have a price list for printing orders given by the rule:
The price of printing q reams is $(-0.6q+10) per ream.
You want to know what size order will give the largest total revenue.
10 To determine the total revenue for
printing 5 reams you multiply 5 by the
price of printing 5 reams. Do you see
that this is the same procedure you follow to calculate the area of the rectangle at q = 5 in the diagram to the right?
Thus the area of the rectangle given by
q is the same as the total revenue of
printing q items. Use this fact to tell
why the total revenue is small for small
quantities, large for “quantities in the
middle,” and small again for quantities
out near 11 and 12. Demonstrate this
fact on the diagram.
10
8
6
4
2
p
.
......
..
.. ......
....... p = −0.6q + 10
..
.......
..
..
..
...........................................•...
........................................................................ .......
.......
........................................................................
.......
........................................................................
.......
........................................................................
........................................................................
.......
........................................................................
.......
........................................................................
.....
........................................................................
........................................................................
........................................................................
........................................................................
........................................................................
.............................................................................................................................................................
q
2
4
6
8
10 12
11 The recipe for the area is q · p, where p is the height of the graph at quantity q. This is also
the recipe for TR from the price list. That is TR = q · p. Use the formula for p to find the
formula for TR in terms of q, and then, using the vertex formula, find the quantity q that
gives the largest value of TR. (It’s OK if you don’t get a whole number.)
The curvy graph will come into play if we are given the graph of average cost. Suppose the average cost
of printing q reams is given by the formula
AC(q) =
8
+ 4.
q
We’d really like to know about total cost. Remember that, if we’re given T C, then we compute AC by
dividing by q:
T C(q)
.
AC(q) =
q
Now, since we’re given AC(q), we’ll need to multiply by q to get T C(q):
T C(q) = q · AC(q).
102
W ORKSHEET #14
12 The total cost of printing q reams is
q times the average cost of printing q
reams. Use this fact to calculate the total cost of printing 4 reams, 6 reams, 8
reams. Then tell why the total cost of q
reams is given by the area of the rectangle at q, using the curvy graph.
13 Whether you are calculating total cost
or area, you use the recipe q · AC(q).
Use this recipe and the formula for
AC(q) to write down a formula for the
area. This formula should look somewhat familiar. It is the total cost formula if paper costs $4 a ream and you
have a fixed cost of $8.
10
8
6
4
2
The Area-of-the-Rectangle Function
p
.
...... ...
Price .. ..
.. ......
........... p = −0.6q + 10
..
..
... .........
..
.... ........
..
...... .......
..
........... ........
..
............................
..
................................
..
8
..
.......
AC(q) = + 4
.. .......
q
..
....
.. AC ..
..
..
..
.............................................................................................................................................................
q
2
4
6
8
10 12
Thus the area-of-the-rectangle function is a graphical representation of calculating TR from a price list and
calculating T C from AC.
14 Now using your formulas for T R and
T C, write down a formula for profit in
terms of q. Use the vertex formula on
this quadratic to find the value of q that
produces the largest profit.
15 Explain in your own words why the
profit can be read from the price list and
average cost graphs to the right as the
shaded area.
10
8
6
4
2
p
..... ..
.. ...
Price
.. ......
..
.............
..
.. ......
..
.. . . . . . . . ........ . . . .............
............................................................................. .......
....................
.........
...
..............Profit
..
.
... . . ...........................................................................
.......................................
..
.......
..
.......
..
..
.......
..
AC ..
..
..
..
.............................................................................................................................................................
q
2
4
6
8
10 12
16 Suppose that costs and your price list have shifted so that the price per ream for a batch of q
reams is given by $(−0.55q + 12) and the average cost of printing q reams is given by 7q + 5.
At what quantity will profit be maximized? (Round to the nearest whole number.) What is
the maximum profit?
W ORKSHEET #14
The Area-of-the-Rectangle Function
103
Note: In exercise #16, you were told to round your quantity to the nearest whole number. However, in
Worksheet #12, you were told to round up to the next higher whole number to maximize profit. How you
round will depend on which method you use:
• If you maximize profit by finding the quantity at which M R = M C, then you must round up to the
next whole number. (Do you remember why? If not, look over Worksheet #12 again.)
• If your profit function is a quadratic and you use the vertex formula to find the quantity that maximizes
profit, then, because of the symmetry of the parabola, you should round (either up or down) to the
nearest whole number.
→
17 We define two new functions, using areas
of rectangles defined by the graphs to the
right.
20
A(x) =
Area of the rectangle determined
by y = f (x) at x,
B(x) =
Area of the rectangle determined
by y = g(x) at x.
15
10
5
a) Write out formulas for A(x) and
B(x).
b) Find the value of A(2) − B(2)
c) Let us define a new function C(x) by
20
C(x) = A(x) − B(x).
15
This new function appears on the
graphs as in the picture to the right.
Write out a formula for C(x). Find
the value of C(2).
10
5
.
...... ..
.. .. f (x) = 12
+6
x
...
..
..
....
..
.
...................................•..........
........
............................................ ..............................
............................................
.... ..................
...........................................................
...................................................•..
.. g(x) = 2x + 1
..................................................................................................................................................................................
1 2 3 4 5
...... .
.. ... f (x)
..
...
..
....
..
..
..
......
............................................•..............
.................................................
............................................
............C(x)
................
....
..................................................•...............
.. .......... g(x)
..................................................................................................................
1 2 3 4 5
d) Find the value of x between 0 and 4 at which C(x) reaches its greatest value.
e) Find the longest interval you can in the interval x = 0 to x = 4 in which C(x) is
decreasing and B(x) is increasing. (Hint: First think about the place where C(x)
peaks.)
f) Find the value of x between 0 and 5 at which f (x) − g(x) = 7.
g) Find the value of x between 0 and 4 at which A(x) reaches its highest value.
104
→
W ORKSHEET #14
18 The graph to the right is of the function
g(r) = √5r + 2. We define the function
A(r) by
A(r) is the area of the rectangle determined
by the point (r, √5r + 2) on the graph.
We consider only those values of r greater
than 1.
The Area-of-the-Rectangle Function
.
......
.. g(r) = √5r + 2
... . . . . . . . . . . . . .......... .....
...........................................................................................................•...... ...........
................
..............................................................................................
...............................................................................................
.................................................................................................................
..............................................................................................
.......................................
...................................A(r)
......................................................................................................................................
..............................................................................................
.................................................................................................................
...............................................................................................
.............................................................................................................................................................................
a) Write out a formula for A(r) in terms of r.
b) Write an equation you could directly solve by using the quadratic formula to answer the
following question: Find the value of r at which the area of the rectangle determined
by the graph is 6.
c) Find the area of the rectangle given by the point on the graph where g(r) = 6.
d) Find the value of r at which the graph of g(r) intersects the straight-line graph f (r) =
4r + 2.
Worksheet #15
The Analysis of Costs III
Introduction
The graphs below are taken from Worksheets #8 and #9. As indicated, they are of the total cost
(T C), variable cost (V C), marginal cost (M C), average cost (AC), and average variable cost
(AV C) of producing bags of potatoes.
4.00
3.20
2.40
1.60
0.80
.
.....
.. ..
.
.
.. ..
.
.
... ...
.
.
.
.
...... ....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...
...........
.
.
.
.
.
.
.
.
.
...
. ... ...
.. .. .. ... ... ... ... ..
..
.
.
..
(Dollars per Bag)
Cost (Hundreds of Dollars)
4.80
p
........
...
...
...
....
..
...
...
..
...
..
...
TC
..
....
...
VC
..
..
..
...
..
...
..
...
.........................................................................................................................................
q
1
2
3
4
5
6
7
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
.p
.......
...
MC
..
...
...
..
Breakeven
...
...
..
...
...
..
AC
...
...
..
...
...
..
.........
...
AV C
•
..
...
...
•..............
..
.....
...
.....
.
...
Shutdown
..
...
...........................................................................................................................................
1
2
3
4
5
6
7 q
...
...
....
...
...
.
...
...
..
.
.
.
.
... ..
.... ..
... .... ........
.
..
... .....
.
..
... ..... .........................................
.
.
... ........
........... ....................
...
............
...
..
....
.
.
....... .....
......
(Hundreds of Bags)
Quantity (Hundreds of Bags)
It happens that these graphs actually come from formulas, so that many of the answers that
we obtained graphically, should be obtainable by algebra. In particular, if we denote the V C of
producing q bags by V C(q) and the M C of producing q bags by M C(q), then
V C(q) =
3
1 3
q − q 2 + q,
30
10
M C(q) =
1 2 3
q − q + 1,
10
5
(where q is given in hundreds of bags, V C(q) is given in hundreds of dollars, and M C(q) is given
in dollars∗ ).
In the course of this worksheet, you will find formulas for the three following quantities.
T C(q) =
AV C(q) =
AC(q) =
∗
Thus, for producing 400 bags, we let q = 4 and find that the V C and M C are:
VC :
MC :
3 · 42
43
−
+ 4 = 1.33 hundred dollars = $133
30
10
42
3
− · 4 + 1 = $0.20
10 5
105
106
W ORKSHEET #15
The Analysis of Costs III
Key Questions
Some of the following can be solved both graphically and algebraically. Some can only be solved
using the graphs.
I. You don’t want your total costs to exceed $320. What is the largest number of bags you
should produce?
II. Bags are to sell at $0.55 each. What is the smallest number of bags you can produce and
have your AV C be at most $0.55?
III. If bags are to sell at $0.55 each, then what is the largest number of bags you can produce and
have variable cost less than, or equal to, total revenue?
IV. Suppose bags all sell for $0.65 each. What is the smallest number of bags you can sell so
that M C is at most $0.65?
Translating the Key Questions into purely algebraic problems is a complex skill that should be developing
for you as you go through these worksheets. However, since not all students have this skill to the same
degree, the first four exercises of this worksheet are written to help you make this translation.
1
You are given formulas for V C and M C.
a) The relationship between V C and T C is that T C is always $80 (=0.8 hundred dollars)
more than V C. Therefore T C(q) = V C(q) + 0.8. Use this fact and the formula for
V C(q) to write the formula for T C(q).
b) The recipe for AV C from V C is that AV C(q) =
for V C(q) to get a formula for AV C(q).
V C(q)
.
q
Use this recipe and the formula
c) The recipe for AC from T C is AC(q) = T C(q)
. Use this recipe and the formula you
q
derived for T C(q) to get a formula for AC(q).
Write these three formulas in the blank spots in the introduction to this worksheet so that all
of your formulas are in the same place.
2
There are now five formulas in this problem. Each belongs to one of the graphs at the
beginning of the worksheet. Clearly mark each of the graphs with the formula that belongs
to it.
Notice that you were not asked to derive the formula for M C from the formula from T C. In theory this is
something you can do from the recipe M C(q) = V C(q + 0.01) − V C(q). (Remember that q is measured
in hundreds of bags. So, 1 bag is 0.01 hundred bags and M C is the change in T C or V C as quantity is
increased by 1 bag.) However, deriving the formula for M C would lead through a morass of algebra, and
even then, the formula we would get would be similar to the one written above, but with some uninteresting
complications. We will deal with these complications (and a quick and easy way of getting the M C formula)
in Math 112.
W ORKSHEET #15
3
The Analysis of Costs III
107
Before trying to translate Key Questions I–IV into algebra, you need to condense the questions. The best way to do this is to think about how you answer these questions using the
graphs.
a) Answer Key Question I by using the T C graph. Do you agree that the question can be
compressed into the following?
I′ . Find the value of q for which T C(q) = 3.20.
b) Answer Key Question II by using the AV C graph. In doing so, complete the following
compressed version of II.
II′ . Find the value of q for which
c) In order to answer Key Question III, observe that if bags sell at $0.55 each, then the T R
graph is a straight line through the origin with slope 0.55. The formula for T R is then
T R(q) = 0.55q. The question asks for the largest value of q for which V C is below
the T R graph. If possible, solve this question graphically; otherwise, explain why you
cannot solve it graphically. Do you agree that the following is a compressed version of
Key Question III?
III′ . For what value of q is V C(q) = 0.55q?
d) Answer Key Question IV by using the M C graph. In doing so, complete the following
compressed version of IV.
IV′ . Find the value of q for which
4
In question 3 we have compressed versions of the Key Questions. In question 1 you wrote
out formulas for the quantities involved. Therefore, you should now be able to translate the
four Key Questions into questions of the form “Find the value of q for which the following
equation is true.” In the four spaces below write down what these equations would be.
I.
II.
III.
IV.
The reward for all this effort should be quick and accurate solutions to the four Key Questions. The
techniques available to us for solving these purely algebraic problems are that we can solve linear equations
and we can apply the quadratic formula and the vertex formula to quadratics. But only two of the above
equations looks like a quadratic (II and IV). This is because the formula for T C (and for V C) has a term that
looks like aq 3 , where a is some number. Because the variable q is raised to the third power, the formulas for
T C and V C are called “third-degree polynomials,” or “cubic polynomials.” There is no easily applicable
“cubic formula” for cubic equations like there is a “quadratic formula” for quadratic equations. Thus, we
can’t solve the equation that comes from Key Question I. Sorry! This should make you appreciate what we
can do.
108
W ORKSHEET #15
The Analysis of Costs III
5
The translation of Key Question II into algebra requires that we find the roots of a quadratic.
Use the quadratic formula to solve it. Check your answer against your graphical solution.
6
The equation for Key Question III is a cubic, but there is a q in every term. Therefore, you
can divide through by q and get a quadratic equation, on which you can use the quadratic
formula. Use this technique to find your answer and check against your graphical answer.
7
Explain why you cannot use this technique of dividing through by q on the equation in Key
Question I.
8
Solve Key question IV by using the quadratic formula.
The important step in this worksheet was translating the wordy questions in English into equations to be
solved. This required that you write out formulas for certain quantities and then use the formulas to write
out equations. Not until you wrote out the equations could you tell whether the problem could actually be
answered numerically. Even then, it required that you do some more or less blind algebraic fiddling in order
to work the equation into one you could solve.
9
Here are some more questions in English about costs and revenue that you should translate
into purely algebraic questions. First do the translation, and then, if you can, solve the
equations you get. If you cannot solve the equation, explain why.
V. What is the lowest value that marginal cost reaches?
VI. Bags are to sell at $0.75 each, so that your marginal revenue is to be $0.75. At what
quantity q will profit be maximized? (Hint: Remember that you want M C and M R to
be equal.)
VII. What is the shutdown price? (Remember that the shutdown price is the smallest value
of AV C and occurs when M C and AV C are equal.)
VIII. What is the breakeven price? (Remember that the breakeven price is the smallest value
of AC and occurs when M C and AC are equal.)
W ORKSHEET #15
→
The Analysis of Costs III
109
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..
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..
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.
TR.. ..
..
..
.......
..
.
.
..
... ....
.
..
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...
... ....
.
..
.
..
... .... TC
.
..
.
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..
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..
.
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..
..........
..
.
.
.
.. ........
. ...
...............................................................................................................................................................................
.
.
10 The graphs to the right are of total revenue
and total cost of producing Framits. The
formula for T C is T C(q) = q 2 + 4q + 4,
where q is in thousands of Framits and T C
is in thousands of dollars.
a) For what value of q will T C be
$80,000?
b) The T R graph is a straight line that
goes through the origin and crosses
the T C graph at q = 7. Write out
the formula for T R as a function of q.
c) What is the smallest number of
Framits you can sell and not lose any
money?
d)
e)
f)
g)
h)
→
q
At what quantity q will you make the greatest profit? What will the greatest profit be?
At what quantity will you make a profit of $6,000?
Your fixed costs increase from $4,000 to $12,000. Write out the new formula for T C.
Under the assumption in (f), at what quantity will you make the greatest profit?
Give a value of fixed costs that would lead to your not making any profit whatsoever,
at any quantity q.
11 You have a straight-line price list for
Blivets, like the one to the right. You
charge $24 each for an order of 1 Blivet
and $18 each for an order of 13 Blivets.
a) Write out a formula that gives the
price p per Blivet for an order of q
Blivets for any q between 0 and 50.
24
18
12
6
p
..... ........
........................
..
........................
..
...............
..
..
..
..
..
..
..
..
..
..
..................................................................................................................................................
q
2 4 6 8 10 12 14
b) Write out a formula in terms of q
for the Total Revenue for a sale of q
Blivets (Recall that TR = p × q.)
c) What is the Marginal Revenue of selling 10 Blivets?
d) The Blivets cost $5 each to manufacture, and you have a Total Fixed Cost of $100.
Write out a formula in terms of q for the Total Cost of producing q Blivets.
e) What quantity of Blivets maximizes Profits? (Assume you cannot sell a fraction of a
Blivet. Your answer should be a whole number of Blivets.)
f) What is the maximum Profit?
Worksheet #16
Distances from Speed Formulas
We have three electronically controlled cars that
roll along three tracks in the same direction. At
time t = 0, the three cars are next to one another.
The graphs to the right tell about the instantaneous speed vs. time for each car.
Instantaneous Speed (ft per minute)
Introduction
7
6
5
4
3
2
1
.
......s
..
......
.
.
.
.
.
.
.
..
.
t ...
..
0.....5.......
+
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2......
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t..)....=......
t...........
(
..
5
.
.
C
.
.
0
............... C .:.....S......
.........
.
.
t...)...=
..................
(
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... ................ A....:.....S...A....
.. ............
..................B....: S
.....
.
.
.
.
.
.
.
..
.
..........B..(t)
..
.........
..........=.. 4
.. ............
...............................................................................................................................................................−
..............0.........4. t
t
1 2 3 4 5 6 7 8 9 10
(Minutes)
Key Questions
I. Recall that A and B are next to one another at time t = 0. Describe how the distance between
the two cars varies as the 10 minutes go by. When are the cars furthest apart?
II. Recall that A and C are next to one another at time t = 0. Describe how the distances
between the two cars vary as the 10 minutes go by. When are they furthest apart?
III. C is going slower than B part of the time, and then faster than B the rest of the time. Is there
a time interval when C is behind B, catches up to B, and then pulls ahead of B? If so, when
does this happen, exactly?
1
Put the three cars in order, from slowest to fastest, at the following times:
(a) t = 1 minute;
(b) t = 3 minutes;
(c) t = 6 minutes;
2
How fast is each car traveling at t = 4 minutes? at t = 5 minutes?
3
Which car, A or B, is further ahead at t = 4.5 minutes?
4
Answer Key Questions I and II.
110
W ORKSHEET #16
Distances from Speed Formulas
111
One of the most straight-forward distance questions we can ask about these cars is:
How far did each car travel in the first minute?
But this is a difficult question to answer. We’ve used the formula:
distance = speed×time
before. And certainly “time” would be 1 minute in this case. But, using car A as an example, the value of
“speed” changes during the interval. Car A’s instantaneous speed at t = 0 is 0 feet per minute and at t = 1
is 0.5 feet per minute. Should we use one of these two speeds in the formula? Or could we use some speed
in between 0 and 0.5 feet per minute? It turns out that, because the instantaneous speed graph is linear,
we can use the speed that is half-way between 0 and 0.5 feet per minute. We find this “half-way” speed by
computing the arithmetic average of 0 and 0.5:
0 + 0.5
= 0.25 feet per minute.
2
This is the average speed of car A over the first 1-minute interval. Notice that the computation of average
speed is different than what we did before when we had information about distance traveled. If you can
determine how far an object travels on a given interval of time, then average speed is simply distance
traveled divided by time elapsed. However, if you don’t know how far the object travels but you do know
that instantaneous speed is linear, then average speed is the speed “half-way” between the instantaneous
speeds at the ends of a time interval.
5
Compute the average speed of car B over the first 1-minute interval by computing the arithmetic average of car B’s instantaneous speeds at the beginning and end of this interval. Do
the same for car C.
If a car travels at an average speed of s feet per minute for t minutes, then the distance the car travels is
d = s × t feet. Over the first minute, car A travels at an average speed of 0.25 feet per minute. So, over the
first minute, the distance car A travels is
d = 0.25 × 1 = 0.25 feet.
6
Compute the distances traveled by car B and car C over the first minute.
7
Find the distance each car travels over the first two minutes by completing the following
steps:
(a) Compute the car’s instantaneous speed at t = 0 and at t = 2.
(b) Compute the car’s average speed over this interval by taking the arithmetic average of
the two instantaneous speeds from part (a).
(c) Use the formula d = s × t to compute the distance traveled.
8
(a) How far does car A travel in the first six minutes?
(b) How far does car B travel in the last four minutes (from t = 6 to t = 10)?
112
W ORKSHEET #16
Distances from Speed Formulas
(c) How far does car C travel from t = 3 to t = 8 minutes?
We’d now like to find formulas for the distance each car travels from time 0 to time t. Since we have three
different cars, we’ll need three different distance functions. We’ll call them
DA (t), DB (t), and DC (t).
We’ll also want to give names to the instantaneous speed function for each car:
SA (t), SB (t), and SC (t).
Notice on the instantaneous speed graphs given at the beginning of the worksheet, we’ve given the formulas
for the instantaneous speeds:
SA (t) = 0.5t;
SB (t) = 4 − 0.4t;
SC (t) = 2 + 0.5t.
We’ll follow the steps of exercise #7 to find the formula for DA (t), the distance car A travels over the first
t minutes. The first step was to find the instantaneous speed at the beginning of the time interval (time 0) and
at the end of the interval (time t). We use the formula for car A’s instantaneous speed to do this. Car A’s
instantaneous speed at time 0 is:
SA (0) = 0 feet per minute.
Car A’s instantaneous speed at time t is simply:
SA (t) = 0.5t feet per minute.
The next step is to find the arithmetic average of these two instantaneous speeds. We add the two instantaneous speeds together and divide by 2:
average speed s =
0 + 0.5t
= 0.25t feet per minute.
2
Finally, to get distance traveled, we use the formula d = s × t. Average speed is s = 0.25t feet per minute
and the length of the time interval is simply t minutes. So, the distance traveled in feet is:
d = 0.25t × t = 0.25t2 .
We have, therefore, a formula for car A’s distance at time t:
DA (t) = 0.25t2 .
9
Follow the same procedure to find formulas for DB (t) and DC (t).
W ORKSHEET #16
Distances from Speed Formulas
113
...... d
50 ...
...
..
10 The formulas for the cars’ distances should
40 ....
all be quadratic functions. Therefore, their
..
..
graphs should be parabolas. Sketch the
30 ...
three distance graphs on the axes at right.
..
..
.
20 ...
..
..
10 ...
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.....................................................................................................................................
m
2
4
6
8
10
11 You answered Key Questions I and II a while ago. Check that these new graphs verify
your responses to those Key Questions. Use your graphs to estimate an answer for Key
Question III.
12 Use the formulas for DB (t) and DC (t) to find the exact answer to Key Question III.
All of our instantaneous speed formulas were linear. We followed the same procedure three times to find
that each distance formula was quadratic. Now suppose that a car has a linear instantaneous speed formula
given by:
S(t) = b + mt.
We’ll follow that same old procedure to find the car’s distance at time t.
13
(a) What is the new car’s instantaneous speed at time 0? at time t?
(b) What is the new car’s average speed from time 0 to time t?
(c) How far does the new car travel from time 0 to time t? Call this distance D(t).
You now have the following: if a car’s instantaneous speed formula is S(t) = b + mt, then its distance
traveled formula is
m
D(t) = bt + t2 .
2
This gives us a shorter procedure for finding a distance traveled formula from a linear instantaneous speed
formula.
For example, if a car’s instantaneous speed is given by s(t) = 3 − 0.75t, then b = 3 and m = −0.75. So,
distance traveled is given by
0.75 2
D(t) = 3t −
t = 3t − 0.375t2 .
2
14 Car J and car K have linear instantaneous speed formulas given by:
SJ (t) = 1 + 0.8t and SK (t) = 5 − 0.6t
(both in feet per minute). At t = 0, the two cars are next to one another. At t = 7, which car
is furthest ahead and by how much?
114
→
W ORKSHEET #16
15 For this problem use the fact that if a car
has an instantaneous speed vs. time function given by S(t) = at + b, then its distance vs. time is given by the function
a
D(t) = t2 + bt.
2
a) The graph to the right shows the instantaneous speed vs. time of the Red
Car. Write out the formula for its instantaneous speed vs. time.
b) Write out the formula for the distance
covered by the car at time t.
2.5
2.0
1.5
1.0
0.5
Distances from Speed Formulas
.
......
..
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......... ....... ......................
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1
2
3
4
c) How far does the car travel from time t = 1 to time t = 5?
d) Write out a formula for the Red Car’s average trip speed at time t.
e) Suppose r is a particular time between 0 and 4. In terms of r, write out a formula for
how far the car travels between time r and 1 minute later. Simplify your answer.
f) The Red Car passes a tree at time t = 0. A Yellow Car is traveling in such a way that
at all times it is 2 miles ahead of the Red Car. Write down a formula for the distance
between the tree and the Yellow Car at time t.
→
.
...... ...
15 .......... ....... .................
.
..
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.
....
..
MC ..........
.....
.
.
.
M R(q) = aq + b
12 ..
..
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.
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....
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.
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.
.... ......
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.
M C(q) = cq + d.
9 .......... ....... ............. ....... ....... ....... ...................................
..
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.
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.
.
..
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.
The formulas for T R and T C are then
.
.
.
.
.
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6 ... ....... .......................
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given by:
.
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3
.
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T R(q) = a2 q 2 + (b − a2 )q
..
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................................................................................
............................................
.
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.
c 2
c
T C(q) = 2 q + (d − 2 )q.
3
6
9 12 15 18
a) The straight line graphs to the right show the actual M C and M R graphs of manufacturing Gizmos. Use the graphs to determine the values of a, b, c, and d.
16 The formulas for M R and M C of manufacturing Gizmos are:
b) Find the profit you make for manufacturing 9 Gizmos.
c) What is the change in profit as you move from q = 6 to q = 8 Gizmos.
d) Write out the formulas for average revenue and average cost of producing q Gizmos.
e) Suppose marginal revenue is decreased by $1 for every quantity q. Write out the new
formula for T R.
→
Distances from Speed Formulas
17 Two toy cars are traveling next to one another along a racetrack. They both pass the
start at time t = 0. At t = 4, both cars
are traveling at a rate of 6 yards per second. The graphs to the right are of the instantaneous speed vs. time for the cars. In
general, if a car has an instantaneous speed
formula s = mt + k, then its distance from
the start is given by d = m2 t2 + kt.
115
12
9
6
3
(yds/sec)
W ORKSHEET #16
.....
.........
.. ....A
.......
..
....... .......................B
..
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.......
..........
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.........................................................................................................
2
4
6
8
(seconds)
a) Write out the distance vs. time formulas for the two cars. Call these functions A(t) and
B(t).
b) At what time in the first 8 seconds are the two cars furthest apart? What is the greatest
distance between them in this interval?
c) Find the distance between the cars when A’s instantaneous speed exceeds B’s instantaneous speed by 2 yds/sec.
d) What is A’s average trip speed at time t = 3 seconds?
e) At what time is B’s average trip speed equal to 6 yds/sec?
f) How far does A travel in the time interval t = 2 to t = 2.5 seconds?
g) Write out a formula for how far B travels from time t = r seconds to t = r +1 seconds.
Simplify your answer.
→
18 The marginal revenue function for producing Twinkles is linear. It is known that the M R of
producing 2 Twinkles is $10, and the M R of producing 22 Twinkles is $0. It is also known
that if an M R function is given by M R(q) = Aq + B, then the T R function is given by
T R(q) = A2 q 2 + (B − A2 )q.
a) Use the given information to determine formulas for the M R and T R of producing q
Twinkles.
b) What is the additional revenue you bring in as you move from producing 4 to 8 Twinkles?
c) The total cost function for producing Twinkles is linear. You have fixed costs of $20
and a total cost of $29 for producing 3 Twinkles. Write out the formula for the profit
you earn in producing q Twinkles.
d) For what value of q is the total revenue of producing q Twinkles largest?
e) For what value of q is the profit of producing q Twinkles largest?
f) Find the interval of values over which profit decreases and total revenue increases.
g) Write out a formula for the average cost of producing q Twinkles.
h) Find a value of q at which the average revenue is $7 per Twinkle.
116
W ORKSHEET #16
Distances from Speed Formulas
Exponential and Logarithmic Functions
117
Worksheet #17
Growth of Sequences
Key Questions
Suppose you want to select a company in which to invest. The following table gives the profits of
four companies over five periods of time.
Profit (in millions of dollars) per period
Period
Company
0
1
2
3
4
5
6
Rutabega Computer Co.
16
20
24
28
32
36
Monkey Pants, Inc.
1.125
2.25
4.5
9
18
36
Aginit Info/Tech
4
27
4
9
4
3
4
12
36
Magic Bus Cruiselines
1
8
15
22
29
36
12
I. Based on the profit data, which company has the greatest profit after period 6 (and, thus,
possibly a better return on your investment)?
II. Which company will have the greatest profit after period 12 (in case this is a longer-term
investment)?
A sequence is a list of numbers in some specified order (not necessarily numerical order). Each of the
following lists is an example of a sequence.
2, 4, 6, 8, 10, ...
0, 1, 0, 1, 0, ...
1, 3, 2, 4, 3, 5, 4, 6, ..
The numbers that make up a sequence are called the terms of the sequence.
1
Give the next two terms in each of the following sequences. Briefly explain how you get
your answers.
A:
40, 50, 60, 70, 80,
,
, ...
B:
64, 71, 78, 85, 92,
,
, ...
C:
2, 6, 18, 54, 162,
,
, ...
D:
100, 50, 25, 12.5,
,
, ...
119
120
W ORKSHEET #17
Growth of Sequences
Sequence A is an example of an additive sequence. To get a new term, we always add 10 to the last
term found. The number 10 is called the increment of this additive sequence. More formally, a sequence is
additive if there exists a number r such that each term in the sequence is obtained by adding r to the term
before it. The number r is the increment of the sequence.
2
Which of the other sequences in exercise #1 is an additive sequence? Give its increment.
How can you tell the others are not additive?
We can use functional notation to denote the terms in a sequence. Specifically,
A(0) = 40, A(1) = 50, A(2) = 60, etc.
Since A is an additive sequence with increment 10, we know that A(k + 1) = A(k) + 10 for any number k.
That is,
A(1) = A(0) + 10;
A(2) = A(1) + 10;
A(3) = A(2) + 10; etc.
3
Which of the following statements about the A-sequence are correct for every value of k?
(a) A(k) = A(k − 1) + 10
(b) A(k) = A(k + 1) − 10
(c) A(k + 2) = A(k) + 20
(d) A(k + 1) = A(k + 10)
(e) A(k + 3) = A(k) + 30
(f) A(k + 1) − A(k) = 10
Item (f) in the previous exercise actually gives a good method to test a sequence to see if it’s additive.
Item (f) says that the difference between any two consecutive terms in the sequence is always equal to the
increment. So, if you have a sequence and you want to know if it’s additive, look at the difference between
consecutive terms. If this difference is always the same number, then the sequence is additive and that
number is the increment.
4
For the B-sequence from exercise #1, fill in the blanks with the correct value.
(a) B(1) − B(0) =
(b) B(2) − B(1) =
(c) B(3) − B(2) =
(d) B(4) − B(3) =
(e) B(k + 1) − B(k) =
W ORKSHEET #17
Growth of Sequences
121
(f) B(k + 1) = B(k)+
(g) B(k) = B(k + 1)−
(h) B(k + 2) = B(k)+
(i) B(k + 3) = B(k)+
The equation A(k + 1) = A(k) + 10 is called a recursive formula for the A sequence. We can use it
to find the value of A(k + 1), but only if we know the value of A(k). What if we wanted to know the value
of A(100)? If all we had was the recursive formula, we’d need to know the value of A(99). But to get the
value of A(99), we’d need the value of A(98), and so forth. What we need is an explicit formula, a formula
that will give us the value of A(k) without of having to know all the terms that came before. We can develop
an explicit formula by noticing a pattern among the terms:
A(0) = 40
A(1) = A(0) + 10 = 40 + 10
A(2) = A(1) + 10 = 40 + 10 + 10
A(3) = A(2) + 10 = 40 + 10 + 10 + 10
and so forth.
We can rewrite each of these equations like so:
A(0) = 40 + 0 × 10
A(1) = 40 + 1 × 10
A(2) = 40 + 2 × 10
A(3) = 40 + 3 × 10
and so forth.
Since we get each new term by adding 10 to the last term, this pattern will continue. So,
A(k) = 40 + 10k.
This is an explicit formula for sequence A. Note that the 40 came from the fact that A(0) = 40 and the 10
is the increment of the sequence.
5
Check that the explicit formula gives the correct values for A(4), A(5), and A(6). (See your
answers to exercise #1.) Give the values of A(100), A(1003), and A(1, 000, 007).
6
Find an explicit formula for sequence B. Use it to find the values of B(100), B(1003), and
B(1, 000, 007).
Let’s get back to sequences C and D from exercise #1. You saw in exercise #2 that these were not additive
sequences. You don’t get from one term to the next by always adding the same number. You get from one
term to the next by always multiplying by the same number.
122
7
W ORKSHEET #17
Growth of Sequences
Fill in the blanks with the correct values.
(a) C(1) =
(b) C(2) =
(c)
C(3)
=
C(2)
(d)
C(k + 1)
=
C(k)
(e) C(k + 1) =
(f) C(5) =
(g) C(5) =
(h) C(4) =
× C(0)
× C(1)
× C(k)
× C(3)
× C(2)
× C(1)
If there exists a number m such that each term in a sequence is obtained by multiplying the term before
by m, then that sequence is a multiplicative sequence and the number m is called the multiplier. To see
if a sequence is multiplicative, divide consecutive terms (like in items (c) and (d) above). If you always get
the same number, then the sequence is multiplicative and that number is the multiplier.
8
Sequences C and D are both multiplicative sequences. What are their multipliers? For each
sequence, give a recursive formula (an equation that relates the k+1st term and the k th term).
We can develop explicit formulas for multiplicative sequences in much the same way we did for additive
sequences. For example, since the multiplier of sequence C is 3, we have:
C(0) = 2
C(1) = C(0) × 3 = 2 × 3
C(2) = C(1) × 3 = 2 × 3 × 3
C(3) = C(2) × 3 = 2 × 3 × 3 × 3
and so forth.
Rewriting, we get:
C(0) = 2 × 30
C(1) = 2 × 31
C(2) = 2 × 32
C(3) = 2 × 33
and so forth.
W ORKSHEET #17
Growth of Sequences
123
This pattern will continue, giving the explicit formula:
C(k) = 2 × 3k .
9
Check that the explicit formula gives the correct values for C(4), C(5), and C(6). (See your
answers to exercise #1.) Find the values of C(10) and C(19). If your calculator can handle
it, find the value of C(99).
10 Find an explicit formula for sequence D. Use it to find the values of D(10) and D(19). If
your calculator can handle it, find the value of D(99).
11 There are sequences that are neither additive nor multiplicative. Explain how you can tell
that the sequence 10, 20, 40, 70, ... is neither additive nor multiplicative.
12 The profits for the four companies in the Key Questions form four sequences. For each,
determine whether the sequence is additive or multiplicative or neither. Find recursive and
explicit formulas for each. Finally, answer the Key Questions.
13 Suppose you start a new job with an annual salary of $40,000. You are promised a 2% raise
at the end of every year. Fill in the following table.
Salary at
Dollar amount
Year
end of year
of raise
0
$40,000
1
$40,800
$800(= 0.02 × 40, 000)
2
(= 0.02 × 40, 800)
3
4
14 Your salaries in successive years form a sequence. Is this sequence additive? If so, what is
its increment? Is this sequence multiplicative? If so, what is its multiplier?
Rather than computing the dollar amount of each raise, we could have thought of your salaries as follows:
after each raise, your new salary is 102% of the old salary. That is, after one year, your salary is $40, 000 ×
1.02 = $40, 800; after two years, your salary is $40, 800 × 1.02 = $41, 616; etc. The annual percentage
change in your salary is 2%, which means that the annual proportionate change in your salary is 0.02,
which means that the multiplier of the salary sequence is 1.02. In general, if a quantity changes by the
same percentage every year, then successive values of that quantity form a multiplicative sequence. The
proportionate change p is the annual percentage change expressed as a decimal (i.e., take the percentage
change and divide by 100 to get p). The multiplier of the sequence is then m = 1 + p.
15 Let S(k) represent your salary at the end of k years.
124
W ORKSHEET #17
Growth of Sequences
(a) Give a recursive formula that relates S(k + 1) to S(k).
(b) Give an explicit formula for S(k) and use it to find your salary at the end of 10 years
(assuming you receive no other raises).
(c) What is the proportionate change in your salary over those ten years? (HINT: If a
quantity changes from an OLD value to a NEW value, then
proportionate change =
NEW − OLD
.
OLD
In this case, your salary is changing from the OLD value of $40,000 to the NEW value
of S(10).)
(d) What is the percentage change in your salary over those ten years?
Summary:
• A sequence A is additive if you always add the same increment r to get from one term to the next. That
is, for any number k,
A(k + 1) = A(k) + r.
This equation is a recursive formula for the sequence. An explicit formula for the term A(k) is:
A(k) = A(0) + rk.
• A sequence M is multiplicative if you always multiply by the same multiplier m to get from one term
to the next. That is, for any number k,
M (k + 1) = M (k) × m.
This equation is a recursive formula for the sequence. An explicit formula for the term M (k) is:
M (k) = M (0) × mk .
• If a quantity changes by the same percentage over a fixed time period, then the successive values of
that quantity form a multiplicative sequence whose multiplier is:
m=1+
→
16
percentage change
= 1 + proportionate change.
100
(a) Suppose A(k) is an additive sequence with increment 5.
i. Give a recursive formula for A(k).
ii. B(k) is a sequence with the following property:
B(0) = 2A(0) + 4, B(1) = 2A(1) + 4, B(2) = 2A(2) + 4, ...
Is B(k) an additive sequence? If so, find its increment. If not, explain why not.
W ORKSHEET #17
Growth of Sequences
125
iii. B(1) = 22. Compute the value of B(108).
(b) Suppose C(k) is a multiplicative sequence with multiplier 9.
i. Give a recursive formula for C(k).
ii. D(k) is a sequence with the following property:
q
q
q
D(0) = 2 C(0), D(1) = 2 C(1), D(2) = 2 C(2), ...
Is D(k) a multiplicative sequence? If so, find its multiplier. If not, explain why
not.
iii. D(2) = 990. Give an explicit formula for D(k).
→
17 We have two sequences of positive numbers: F (0), F (1), F (2), . . . and G(0), G(1), G(2), . . ..
The two sequences have the property that: The k th term of the G-sequence is equal to the
(3k)th term of the F -sequence. This can be restated as G(k) = F (3k), or, equivalently,
F (0), F (1), F (2), F (3), F (4), F (5), F (6), F (7),
k
k
G(0)
k
G(1)
G(2)
The multiplier of the F -sequence is m (where m > 1).
The multiplier of the G-sequence is n.
a) Write out an equation that relates the letters m and n.
b) In terms of the letter m, what number must you multiply G(4) by to get G(6)?
c) In terms of the letter n, what must you multiply F (6) by to get F (8)?
d) Let j be an integer. In terms of the letters m and j, tell what you must multiply F (7)
by to get F (7 + j).
e) Put the symbol <, >, or = in the boxes below:
→
F (16)/F (15)
F (17)/F (16)
F (16) − F (15)
F (17) − F (16)
F (16)/F (12)
F (6)/F (2)
18 Suppose we have a function y = f (t) = K · (1.4)t . The value of K varies in each of the
questions below. However, in each case, you can either deduce the value of K, or you don’t
need it, so that your answers should not involve K.
a) Suppose f (2) = 19. What is the change in the value of y as t goes from 2 to 2.5?
b) Suppose f (2) = 100. Find f (0) and f (3).
126
W ORKSHEET #17
Growth of Sequences
c) Suppose f (0) = 10. Find f (5) − f (4.5).
d) By what factor is f (2) multiplied to get f (5)?
e) What is the percentage change in the value of y as t goes from 7 to 10?
f) What is the proportionate change in y as t goes from 3 to 3.75?
Worksheet #18
The Inheritance
Key Question
Your old Uncle Charlie, the financier, has just died and left you a legacy. However, he has laid out
three different conditions under which you may receive your money, and you must now choose
which one you want. Read the three conditions below, and then tell which you think gives you the
greatest amount of money.
(A) An account is set up in your name and $10,000 is deposited in it. An additional $10,000 is
deposited in the account at the end of each of the next 25 years. You are given the balance at
the end of the 25th year, after the annual deposit has been made.
(B) An account is set up in your name and $10,000 is deposited in it. At the end of each year, 15%
of the current balance of the account is credited to the account. You are given the balance at
the end of the 25th year, after the interest has been credited.
(C) This condition is the same as (B) except that your account is credited with 7.5% of the
balance every six months.
We will abbreviate the balance at the end of k years by A(k), B(k) and C(k), respectively. The
Key Question is, then: Which is bigger A(25), B(25) or C(25)?
1
We have A(0) = 10, 000, A(1) = 20, 000, A(2) = 30, 000, while B(0) = 10, 000, B(1) =
11, 500, and B(2) = 13, 225. Compute the exact values of A(3), A(4), B(3), and B(4).
2
One way to think about these problems is to arrange things along a time-line, as I have done
for the Condition (A) below
Number of Years
0
1
3
2
(A)
Amount
10,000
20,000
30,000
+10, 000
40,000
...
...
.....
.......
........
...
....
....
............
....
.....
......
......................
...
....
....
............
....
.....
......
......................
...
....
....
............
....
.....
......
......................
+10, 000
+10, 000
···
···
Write out balances along a time-line for an account following Condition (B).
Number of Years
0
1
3
2
(B)
Amount
10,000
11,500
...
....
....
.............
....
.... .
......
.....................
...
....
....
.............
....
.... .
......
.....................
...
....
....
.............
....
.... .
......
.....................
×1.15
×1.15
×1.15
127
...
...
.....
.......
.......
···
···
128
W ORKSHEET #18
The Inheritance
3
Condition (A) leads to an additive sequence. What is its increment? Condition (B) leads to
a multiplicative sequence. What is its multiplier? What is its proportionate change?
4
Give explicit formulas for A(k) and B(k).
5
Now you should be able to choose between Conditions (A) and (B) by computing A(25) and
B(25).
6
It didn’t even look very close in the early years, but Condition (B) actually comes on strong
and eventually catches up to Condition (A). Explain why.
7
Suppose there had been a misunderstanding and Condition (B) was really to be at 13%
interest, but to run for 30 years. Do you agree that the balances would be a multiplicative
sequence with multiplier 1.13 (and proportionate change 0.13)? Give an explicit formula for
B(k) under this new condition and compute B(30).
We have backed into a famous formula in finance. If you start with P dollars, which grows by proportionate increases of p every base period (like a year), then you have a multiplicative sequence with a time line
that looks like
Period
0
Amount
P
1
P (1 + p)2
P (1 + p)
(k − 1)
3
2
P (1 + p)3
...
...
....
.............
....
.... .
......
.....................
...
...
....
.............
....
.... .
......
.....................
...
...
....
.............
....
.... .
......
.....................
×(1 + p)
×(1 + p)
×(1 + p)
...
...
.....
.......
........
···
k
P (1 + p)(k−1)
···
P (1 + p)k
...
...
....
.............
....
.... .
......
.....................
Thus, at the k th period over, P dollars are worth an Amount A which can be written as A = P (1 + p)k .
But then there are two slightly irritating confusions:
(i) You will often be given the percent change rather than the proportionate change. But the proportionate
change is simply the percentage, written as a decimal. We might write the percent change as (r ×
100)%, in which case, the proportionate change is simply r.
(ii) Interest is not generally compounded once a year, as in Condition (A), so that to use the formula you
have to talk about the number of compounding periods—where a compounding period may be a year,
a month, a day, or any other subdivision of a year.
Then, in the standard form, we get
Compound Amount Formula (CAF)
A = P (1 + r)N , where
P = principal = amount initially deposited
r = interest rate, expressed as a decimal
N = number of times compounding has taken place
A = amount in account after N compoundings
W ORKSHEET #18
The Inheritance
129
8
Under the Condition (B) there was to be a $10,000 principal, the interest rate was to be 15%
(r = 0.15) and the money was to be compounded 25 times. Use the CAF to find the final
amount of the inheritence in condition (B). Compare to your answer to exercise #5.
9
Use the CAF to find the final amount if the interest rate is 13% and the term is 30 years.
Check against your answer to exercise #7.
10 To deal with Condition (C) of the Key Question, we can either get into subtleties of the CAF,
or we can go back to the time-line.
k
midyear
k+1
C(k)
X
C(k + 1)
The bank adds 7.5% interest to the balance C(k) at midyear. So the balance then is X =
(1.075)C(k), right? Then they add 7.5% to this new balance, X, at the end of the year, to
get C(k + 1). So, to get C(k + 1), they multiply X by 1.075, right? Put these facts together
to find out what number they multiply C(k) by to get C(k + 1).
11 The work in exercise #10 says that we can treat Condition (C) exactly like Condition (B),
except with an annual interest rate of 15.5625%. Use this fact and the CAF to find the value
of your inheritance at the end of 25 years under Condition (C).
12 Now go back and get a clear answer to the Key Question: which of the three conditions
described at the beginning of this worksheet will give you the largest inheritance?
Worksheet #19
Going Backwards in Time
Key Questions
I. Your first child was born today, and you want to put enough money into the bank today so
that she will have enough money to go to college in 18 years. You estimate that it will cost
$40,000 to go to college then. You have located a bank that will guarantee interest of 8% per
year over the next 18 years. How much money should you put into the bank today to be able
to give her the $40,000 in 18 years?
II. You work in a biology lab, and you grow a bacteria that doubles its population every 20
minutes. You want to have a bacteria colony that has 2,048 million bacteria 200 minutes
from now. How many bacteria should it have 60 minutes from now?
We will adopt many of the notations and techniques of Worksheets #17 and #18 to deal with these Key
Questions. That is, we will use the shorthand:
1
A(k)
=
The balance in your daughter’s account at the
end of the k th year (right after the bank has compounded it for the k th time).
B(k)
=
The population of the bacteria colony after k
twenty-minute time periods have elapsed.
Translate the key questions into functional notation.
We will use time-lines as in Worksheet #18 to work with the two sequences A(0), A(1), A(2), . . . and
B(0), B(1), B(2), . . ..
The time-line below shows the population of the bacteria colony at 20-minute intervals. The quantities
will be millions of bacteria.
Time
Population
0
1
2
B(0)
B(1)
B(2)
3
···
B(3) · · ·
8
B(8)
9
10
B(9)
1,024
B(10)
2,048
To go one unit from left to right on this time line you double the population. Thus, to go from B(9) to
B(10) you multiply by 2.
2
(a) What must you do to get the number B(9) from the number B(10)?
(b) What must you do to the number B(9) to get the number B(8)?
130
W ORKSHEET #19
Going Backwards in Time
131
(c) What must you do to get the number B(8) from the number B(10)?
(d) What must you do to get the number B(6) from the number B(10)?
(e) What must you do to get the number B(3) from the number B(10)?
3
Answer Key Question II.
4
Imitate the process used in the previous questions to find how many bacteria you must start
with in your colony so that you will have 2,048 million bacteria at the end of 200 minutes.
5
Biologists can control the rate of growth of a bacteria colony by changing the chemicals in
which the colony is growing. Suppose you change the chemicals so that the bacteria colony
is multiplied in size only by a factor of 1.5 every 20 minutes. Then, the pattern along the
time line is as follows:
×(1.5)2
×1.5
×1.5
B(k + 1)
B(k)
B(k + 2)
÷1.5
÷1.5
÷(1.5)2
What must you do to go, in one leap, from B(8) to B(10)? What must you do to go, in one
leap, from B(10) to B(7)?
6
What must you do to go, in one leap, from B(10) to B(3) with these new chemicals? How
many bacteria should you have in your colony at 60 minutes in order to have 2,048 million
bacteria in your colony at 200 minutes?
Next we would like to apply the same method of analysis used on the bacteria to find out how much
money you should put in your daughter’s account today. But before doing so, we should remember the
relations between percentage interest and the multiplier of a sequence. Recall that 8% = 0.08, so that the
proportionate change in this account is 0.08 and the multiplier of the sequence is 1.08. That is, the bank
multiplies the balance by 1.08.
In general, if interest rate = (r × 100)%, then the proportionate change is r and the multiplier is 1 + r.
Conversely, if the multiplier is m, then the proportionate change is m−1 and the interest rate is (m−1)×100
percent.
7
The multiplier in your daughter’s account is 1.08. Thus a time-line representation of the
balances looks like this:
132
W ORKSHEET #19
Going Backwards in Time
×(1.08)2
×1.08
×1.08
A(k)
A(k + 1)
A(k + 2)
÷1.08
÷1.08
÷(1.08)2
If your daughter has $40,000 in her account on her 18th birthday, how much was in her
account on her 16th birthday? On her 14th birthday? (Try to go in “one leap”.)
8
Complete the following one-leap-scheme, by deciding how much you must multiply or divide by to go between A(0) and A(18).
×(1.08)(
A(0)
)
..................................................................................... ...
...............
.................
.... ....
.............
....................
.
.............
... ..................
...................................................................................
÷(1.08)(
A(18)
)
Use this scheme to find out how much money you should put in your daughter’s account the
day she is born in order to have $40,000 on her 18th birthday—if the money is compounded
8% annually. You have then answered Key Question I.
The problem we have just worked on is considered an extremely important one in economics, accounting
and finance. Money grows with time in a definite pattern, and you ask: “How much must I have now in
order to have a given amount later?” Thus you need $10,009.96 in your daughter’s account at her birth
in order to have $40,000 there 18 years later. Financial types say that $10,009.96 is the present value of
$40,000, 18 years from now. They also say that $40,000 is the future value of $10,009.96, 18 years from
now.
9
Suppose you decided that your daughter needed $50,000 at age 20, but you located a bank
that paid 10% interest per year. How much should you put into her account at birth? i.e.,
What is the present value of $50,000 at 10% in 20 years?
Next we want to write formulas that take the place of the time-line schemes we have been using. Suppose
you have a bacteria colony which starts out with N bacteria and it has a multiplier of m every period. Then
you have
×m
..........................................
.
.......
×m
..........................................
.
.......
B(1)
N
..............
.
..................................
÷m
×m
..........................................
.
.......
B(2)
..............
.
..................................
÷m
×m
B(3)
..............
.
..................................
÷m
×m
..........................................
.
.......
..............
.
..................................
÷m
..........................................
.
.......
···
×m
..........................................
.
.......
B(k − 1)
..............
.
..................................
÷m
B(k)
..............
.
..................................
÷m
W ORKSHEET #19
Going Backwards in Time
133
so that B(k) = N · mk .
Likewise, if you have a bank account which starts with P dollars and earns interest at a rate of (r × 100)
percent, then your account has a multiplier of (1 + r) and by a time-line scheme identical to the one above,
you get A(k) = P (1 + r)k , which is your old friend, the Compound Amount Formula.
10 Key Question I asked what P should be when k = 18, r = 0.08 and A(18) = $40, 000. So
you must solve the equation 40, 000 = P · (1.08)18 . Solve this equation for P and compare
the answer you get this way to the answer to exercise #8.
11 Use the CAF to find the present value of $50,000, 20 years from now, if it is compounded at
an annual rate of 10%. Compare your answer with your answer to exercise #9.
12 For the bacteria colony in Key Question I, use the formula B(k) = N · mk to determine
the present population of 2,048 million bacteria, 200 minutes from now. Compare to your
answer to exercise #4.
The moral is that to find a present value you divide by a power of the multiplier and to find a future value
you multiply by a power of the multiplier.
→
13
a) Find the present value of $100 ten years from now, to be compounded at an annual rate
of 12%.
b) Bacteria cells multiply at a rate of 75% every half-hour (i.e., have a multiplier of 1.75).
How many bacteria must you have now in order to have 100 million 3 hours from now?
c) If you put $100 in a bank account compounded at an annual rate of 15%, how much
will you have after 8 years?
d) Money is to be compounded at an annual rate of 12%. How much should be in an
account at the end of 4 years, if it is to have $10,000 at the end of 11 years?
e) Find the present value of $500, 10 years from now, to earn interest at 14% annually.
f) You want 500 million bacteria 6 hours from now. They double every half-hour. How
much should you start with?
g) Find the present value of $600, 7 years from now, to earn interest at 8% per year.
h) What is the present value of $500, six years from now, if the balance of an account is
increased by 25% every two years?
→
14 Mary and Charlie both have bank accounts. Mary’s earns 8% per year and Charlie’s earns
7% per year.
a) If Mary puts P dollars in her account now, how much must Charlie put in his in order
for them to have identical balances in 4 years?
134
W ORKSHEET #19
Going Backwards in Time
b) If Mary puts P dollars into her account now, how much should Charlie put into his
account so that at the end of the 6th year, his balance will be identical to the balance
Mary has at the end of the 5th year?
c) Suppose that Mary and Charlie put P dollars in each of their accounts now, but then in 4
years they switch, with Mary putting her money into what used to be Charlie’s account
and Charlie putting his money into what used to be Mary’s account. Is it possible to
tell who would have more money 8 years from now? Is it possible to tell who would
have more money 7 years from now? Explain.
Worksheet #20
Partial Compounding Periods
Key Questions
I. XYZ Corporation issues promissory notes in $1,000 denominations under the following
terms. You give them $1,000 now, and eight years later they will give $2,000 back to you.
You bought $8,000 worth of these notes 4 years ago and wish to sell them today to a friend
who will cash them in for $16,000 4 years from now. How much money should your friend
give you to make the deal fair?
II. You have a bacteria colony that triples its population every hour. You start with 10 million
bacteria and come back 20 minutes later and then 20 minutes after that. How many bacteria
should you expect to see at each of these two times?
We’ll start with Key Question I.
1
Your friend says that, since the investment will be earning $8,000 in interest over the next
8 years, you’re earning $1,000 in interest each year.
(a) With this approach, fill in the following table:
Time
Supposed Value of Investment
4 years ago
$8,000
3 years ago
2 years ago
1 year ago
$11,000
now
$12,000
1 year from now
2 years from now
3 years from now
4 years from now
$16,000
(b) With this approach, you will pay $8,000 and receive $4,000 in interest in four years.
Your friend will pay $12,000 and receive $4,000 in interest in four years. Is this deal
fair to both you and your friend?
2
You say that, since the investment will be earning 100% interest over the next 8 years and
100 ÷ 8 = 12.5, you’re earning 12.5% interest each year.
(a) With this approach, fill in the following table:
135
136
W ORKSHEET #20
Partial Compounding Periods
Time
Supposed Value of Investment
4 years ago
$8,000
3 years ago
$8,000+8,000(0.125)=$9,000
2 years ago
$9,000+9,000(0.125)=$10,125
1 year ago
now
$12,814
1 year from now
2 years from now
3 years from now
4 years from now
(b) The supposed values of the investment in the second column of the table form a multiplicative sequence. Give the values of the multiplier and the proportionate change.
(See Worksheet 18, #20, for the relationship between the multiplier and proportionate
change.)
(c) Use the last entry in the table to explain why you were wrong to assume that 100% over
8 years was the same as 12.5% per year for 8 years.
3
Your Math 111 TA tells you that the way to make the deal fair is to make sure that you both
earn in interest the same percentage of your respective investments. Let p be that percentage,
expressed as a decimal. Suppose the amount your friend will pay you is $X.
(a) You will pay $8,000 and receive $X from your friend. The amount X should include
your original investment plus (p × 100)% interest. That is,
X = 8, 000 + 8, 000p
OR X = 8, 000(1 + p).
Similarly, your friend will pay $X and receive $16,000. But this $16,000 will also be
your friend’s original investment plus (p × 100)% interest. That is,
16, 000 = X + Xp
OR
16, 000 = X(1 + p).
This gives two equations in two unknowns. Find the values of X and p.
(b) How much will you earn in interest (in dollars) with this approach? How much will
your friend earn?
(c) Does this deal seem fair to both you and your friend?
4
We know from #2 that 100% over 8 years does not mean 12.5% per year for 8 years. But
there is what’s called an effective annual interest rate, a rate that, if applied each year for
8 years, would result in a 100% gain in the investment. Let r be this effective annual interest
rate, expressed as a decimal, and let A(k) be the value of the investment k years after the
notes were purchased. Then, A(0) = 8, 000 and A(8) = 16, 000, for example.
W ORKSHEET #20
Partial Compounding Periods
137
(a) Write an equation that relates A(1) to A(0). (Your equation should contain an r.)
(b) Write an equation that relates A(2) to A(1). Then use your answer to part (a) to write
an equation that relates A(2) to A(0).
(c) Write an equation that relates A(3) to A(2). Then use your answer to part (b) to write
an equation that relates A(3) to A(0).
(d) Your answers to parts (a), (b), and (c) should be revealing a pattern: the numbers A(k)
form a multiplicative sequence. What are its multiplier and proportionate change?
(e) Write an equation that relates A(k) to A(0).
(f) We know that A(0) = 8, 000 and A(8) = 16, 000. Use your formula from part (e)
to determine the value of r, the effective annual interest rate. Express this rate as a
percentage.
5
Suppose you place $1,000 into a bank account that earns 6% interest annually. After 6 months,
you withdraw the entire balance, which is your $1,000 plus some interest. In the following
problem, you’ll determine this balance.
(a) Let A(k) be the balance after k years. What is A(0)? If you left the money in the
account, what would be the value of A(1)?
(b) The 6% annual interest corresponds to some rate of interest r (expressed as a decimal)
per six months. Use your answers to part (a) to show that r does not equal 3% (6%÷2).
(HINT: Show what A(1) would be if you earned 3% every six months for one year.)
(c) Let X be the balance in the account after 6 months. Then, X = A(0) + A(0)r OR
X = A(0)(1 + r) and A(1) = X + Xr OR A(1) = X(1 + r). Combining these, we
get that A(1) = A(0)(1 + r)2 . Use your answers to part (a) to solve for r.
(d) √
If you did part (c) correctly, you should have passed through a step that said that 1+r =
1
1.06 or 1 + r = (1.06) 2 . Recall that, if X is the balance after six months, then
1
X = A(0)(1 + r). This can be re-written as X = A(0)(1.06) 2 . Compute the value of
X, the balance in the account after 6 months ( 12 a year).
In Worksheet #18, we introduced the Compound Amount Formula:
A(k) = P (1 + r)k .
• P =principal
• r =interest rate, expressed as a decimal
• k =number of times compounding has taken place
• A(k) =balance after k compoundings
138
W ORKSHEET #20
Partial Compounding Periods
One limitation with this formula, as derived in Worksheet #18, was that k must be a whole number. We’re
now ready to extend the CAF to allow for fractional values of k. In the last problem, you showed that you
can compute the balance in the account after 6 months ( 21 a year) by using the CAF:
A
1
2
1
= 1, 000(1 + 0.06) 2 .
That is, instead of thinking of k as the number of compoundings that have occurred, we can think of k as the
number of compounding periods (years in this case) that have passed (and this number need not be a whole
number).
6
Suppose you deposit $17,000 into an account that offers 5.5% annual interest. Use the CAF
to compute the balance in the account after 6.25 years.
7
The same types of computations could be done to answer Key Question II.
(a) The population triples every hour. So, we can think of the “compounding period” as
being one hour. How many compounding periods have passed after 20 minutes? After
40 minutes?
(b) Let B(k) be the number of bacteria present after k hours. Write an equation that relates:
i.
ii.
iii.
iv.
B(1) to B(0)
B(2) to B(0)
B(3) to B(0)
B(k) to B(0).
(c) Your answer to part (b)iv gives a formula for B(k) that you should believe works as
long as k is a whole number of hours. However, you could do work similar to what
we did with the bank accounts to show that this formula also works if k is not a whole
number. Use this fact to answer Key Question II.
→
8
(a) Find the value of $10,000 after 4.5 years, if it carries interest of 15% per year.
(b) A bacteria strain multiplies by a factor of 1.5 every hour. How many bacteria are in a
colony after 210 minutes, if the colony started with 5 million bacteria?
(c) Money gains interest at a rate of 50% every 4 years. What is the value of $5,000 after
6 months? after 7 years?
(d) An account gains interest at a rate of 10% per annum. How much must you have now
in order to have $4,000 2.5 years from now?
(e) Bacteria double every 30 minutes. How much must you have now in order to have
5 million bacteria in 45 minutes?
W ORKSHEET #20
→
9
Partial Compounding Periods
139
A bacteria colony increases its population at the rate of 75% every 30 minutes.
a) What is the population 3 hours from now of a colony that has 5 million bacteria now?
b) What would the population have to be now, if the population is to be 27 million 6 hours
and 15 minutes from now?
c) What is the change in population over the next 2 hours if the colony has 7 million
bacteria now?
d) What is the percentage change in the population over any 2 hour period?
e) What is the proportionate change in the population over any 1.5 hour period?
f) By what factor is the population multiplied from now until 2.5 hours from now?
→
10 You have a bacteria colony whose population at time t hours is given by the formula P (t) =
k · mt , where k and m are constants.
a) Express the population of the colony at t = 5 hours in terms of the constants k and m.
Do the same for the population at t = 5.5 hours.
b) Find an expression for the proportionate change in the population from t = 5 to t =
5.5 hours. Simplify as much as possible.
c) The population increases by 90% from t = 5 to t = 5.5. Find the value of the constant m.
d) What is the percentage increase over the time period t = 3 to t = 5.5 hours?
e) Suppose you are now told that the population of the colony at t = 3 hours is 7 million.
Compute the value of the constant k.
f) You have another bacteria colony of the same type as the one just described. Its population is 90% larger. What is the population of this second colony at t = 4 hours?
Worksheet #21
Interest
Key Question
When you invest in a certificate of deposit (CD), you agree to let a financial instituition hold your
money for a pre-determined period of time. In exchange for the privilege of using your money, the
bank pays you interest at an annual rate that depends on the size of your deposit (your principal)
and the time period you choose (the term). In addition, an institution can choose from several
different methods of computing your interest.
Each of the following CDs was advertised on-line at various U.S. Banks during August 2004.
Account
Principal
Term
Annual Interest Rate
Type of Interest
I
$2500
90 days
1.00%
simple interest
II
$1500
54 months
3.45%
compounded annually
III
$500
18 months
1.74%
compounded quarterly
IV
$5000
29 months
2.47%
compounded monthly
V
$500
12 months
1.54%
compounded daily
VI
$1000
30 months
3.11%
compounded continuously
How much interest has each CD earned at the end of its term?
With simple interest, you earn interest only on the principal. You do not earn interest on any interest
you’ve already received. So, for example, with a principal of $4000 earning simple interest of 1.50% annually, you’ll earn 0.015 × 4000 = $60 per year in interest. You will earn $60 in interest every year, no matter
how large the account grows.
principal =
A(0) = $4000
amount after 1 year =
A(1) = $4060
amount after 2 years =
A(2) = $4120
amount after 3 years =
A(3) = $4180
and so forth
1
(a) The values A(0), A(1), A(2), A(3), ... form a sequence. Is it additive, multiplicative,
or neither?
(b) Give an explicit formula for A(t), the amount in the account after t years.
2
Suppose you have a CD offering simple interest at an annual rate of (r ×100)%. You deposit
a principal of $P .
140
W ORKSHEET #21
Interest
141
(a) In terms of P and r, how much interest (in dollars) will you earn each year?
(b) Let A(t) be the amount in the account after t years. In terms of P and r, what is A(0)?
A(1)? A(2)? A(3)?
(c) Give a formula in terms of P , r, and t for the amount in the account after t years.
3
From exercise #2, we have the simple interest formula
A(t) = P (1 + rt)
for the amount in the account after t years. The following questions refer to account (I) in
the Key Question.
(a) What are the values of P and r?
(b) Give a formula in terms of t for A(t), the amount in the account after t years. (NOTE:
Just like in Worksheet #20, this formula works even if t is not a whole number.)
(c) Use your formula to compute the value of the CD at the end of the 90-day term. (TIP:
You’ll need to convert 90 days into a number of years. Banks often use 360 as the
number of days in a year because it’s oh-so-divisible.)
With simple interest, no matter how much interest you’ve earned, the account will only pay a percentage
of the principal. However, most accounts pay a percentage of the entire balance, the principal plus all the
interest already earned. Interest paid on interest is said to be compounded.
4
In Key Question Account (II), you can earn 3.45% interest, compounded annually (once a
year). Letting A(t) be the amount in the account after t years, we have A(0) = 1500.
(a) During the first year, you’ll add 3.45% of your principal to your balance. Compute
A(1), your balance after one year.
(b) During the second year, you’ll add 3.45% of A(1) to your balance. Compute A(2),
your balance after two years.
(c) Give a formula, in terms of t, for your balance after t years. (Again, this formula will
work even if t is not a whole number.)
(d) Use your formula to compute the value of the CD at the end of its 54-month term.
In Key Question Account (III), you earn 1.74%, compounded quarterly. The bank divides the annual
= 0.00435) and credits you with 0.435% of the current balance four times a year.
interest rate by 4 ( 0.0174
4
principal =
A(0)
= $500
amount after one quarter =
A(0.25)
= $500(1.00435) = $502.18
amount after two quarters =
A(0.5)
= $500(1.00435)2 = $504.36
amount after three quarters =
A(0.75)
= $500(1.00435)3 = $506.55
A(1)
= $500(1.00435)4 = $508.75
amount after four quarters (one year) =
142
5
W ORKSHEET #21
Interest
(a) In Key Question Account (III), what would the value of the CD be after 2 years?
3 years? t years?
(b) In Key Question Account (III), what is the value of the CD at the end of the 18-month
term?
Now consider a general account with the following parameters:
P=
A(0) = amount initially deposited
r=
interest rate, expressed as a decimal
n=
number of times per year interest is compounded
The bank takes the interest rate and divides by n. They then credit your account with
balance, n times a year. Your balance after t years is given by the formula
r
A(t) = P 1 +
n
nt
r
n
× 100 % of the
.
(As always, t need not be a whole number — but it must be in years.)
6
In Key Question Account (III), P = $500, r = 0.0174, and n = 4. Check that this formula
gives the same result you computed in exercise #5.
7
In Key Question Account (IV):
(a) What are P , r, and n?
(b) Give a formula for A(t).
(c) Find the value of the CD at the end of the 29-month term.
8
In Key Question Account (V), we’ll use n = 365.
(a) What are P and r?
(b) Give a formula for A(t).
(c) Find the value of the CD at the end of the 12-month term.
All that’s left for us to do is to figure out what a bank means when it advertises interest compounded
continuously. To illustrate this principle, consider an account that pays interest at an annual rate of 12%.
You intend to deposit $250 in this account.
9
You’d like to have $1000 in this account after ten years (without having to deposit any more
money).
(a) If interest is compounded annually, will you reach your goal?
(b) If interest is compounded semi-annually (twice a year), will you reach your goal?
W ORKSHEET #21
Interest
143
(c) If interest is compounded quarterly, will you reach your goal?
(d) Without doing any more calculations, make a guess as to a value of n that will allow
you to reach your goal of a $1000 balance in ten years. (There’s no right answer here;
you’re just guessing.) Choose your guess from the following:
i.
ii.
iii.
iv.
v.
vi.
n = 12 (interest is compounded monthly)
n = 365 (interest is compounded daily)
n = 8760 (interest is compounded every hour)
n = 525, 600 (interest is compounded every minute)
some n larger than 525,600
It’s impossible — at 12% interest per year, you can’t reach $1000 in ten years with
interest alone.
Now I’ll do the computations so we can answer part (d) for real.
Compounding
Formula for A(t)
annually: n = 1
A(t) = 250(1 + 0.12)t
semi-annually: n = 2
A(t) = 250 1 +
quarterly: n = 4
A(t) = 250 1 +
monthly: n = 12
A(t) = 250 1 +
daily: n = 365
A(t) = 250 1 +
every hour: n = 8760
A(t) = 250 1 +
every minute: n = 525, 600
A(t) = 250 1 +
A(10)= balance after 10 years
0.12
2
0.12
4
0.12
12
0.12
365
0.12
8760
0.12
525600
$776.46
2t
$801.78
4t
$815.51
12t
$825.10
365t
$829.87
8760t
$830.02
525600t
$830.03
You can see that the balance increases as n increases; but the larger n gets, the smaller the changes in
the balance.
10
(a) Fill in the following table to demonstrate how much the balance increases as n increases.
change in n
change in balance
from n = 1 to n = 2
$25.32
from n = 2 to n = 4
from n = 4 to n = 12
from n = 12 to n = 365
from n = 365 to n = 8760
from n = 8760 to n = 525, 600
144
W ORKSHEET #21
Interest
(b) Without doing any calculations, estimate how much the balance would change if you
compounded every second instead of every minute.
It turns out that, for this account (P = $250, r = 0.12), the maximum possible balance after ten years is
$830.03. (So, you will not reach your goal of $1000 in ten years.)
If a bank compounds so that, at any given time, your account contains its maximum possible balance,
then they are compounding the interest continuously. In the above example, the maximum possible balance
after ten years was the same as the balance when n = 525, 600. But it’s possible, with a different principal
amount and/or a different interest rate, that the maximum balance could coincide with a different value of
n. How do you tell what the maximum balance is? We have a special formula for interest that is compounded continuously. (Understanding how we come up with this formula requires the concept of a limit
from Calculus.) The formula is:
A(t) = P ert .
Here, A(t), P , and r are all as before:
A(t)=
P=
r=
balance after t years
principal
interest rate, expressed as a decimal
The number e is a special constant, kind of like π: π = 3.14159... and e = 2.71828.... (Your calculator
should have an ex button. Check that you know how to use it: e1 = 2.718281828, e2 = 7.389056099,
e10 = 22026.46579.)
11 In Key Question Account (VI), the interest is compounded continuously. So, we’ll use the
formula
A(t) = P ert .
(a) What are the values of P and r?
(b) Give a formula in terms of t for the amount in this account after t years.
(c) Compute the value of the CD at the end of its 30-month term.
You’ve developed three formulas in this worksheet.
• Simple interest:
• Discrete compound interest:
A(t) = P (1 + rt)
A(t) = P 1 +
• Continuous compound interest:
r
n
A(t) = P ert
nt
W ORKSHEET #21
Interest
145
In each, P is the principal (or the present value of the investment); r is the annual interest rate, expressed
as a decimal; and A(t) is the amount in the account after t years (or a future value of the investment). In
the discrete compound interest formula, n is the number of times per year interest is compounded.
→
12
a) What is the value of $10,000, 8 years from now, if it is placed in an account with a
nominal annual rate of 11% and compounded quarterly?
b) Money is compounded continuously at 9% per year. By what factor is $10,000 multiplied after 2 years?
c) Interest is compounded daily in an account that has an annual rate of 6%. What is the
percentage change of the balance over 3 months?
d) An account bears interest of 8% annually, compounded monthly. What is the present
value of a dollar 2 years from now?
→
13 You are considering putting money into one of three accounts. The description of the ways
in which interest is accrued in the three accounts is as follows:
Account A: 8% annual interest compounded semi-annually
Account B: 8% annual interest compounded quarterly
Account C: 8% annual interest compounded continuously
Compute the following:
a) The change in value of an initial investment of $12,000 in Account A from 3 to 4 years.
b) The percentage change over 10 months in $9,000 put in Account C.
c) The value of a dollar 16 months after it has been placed in Account B.
d) The balance 7 years from now of $400 placed in Account A 5 years from now.
e) The proportionate change of an initial investment of $1,000 placed in Account B from
t = 3 to t = 3.5 years.
f) The present value of $500 to be received from Account C 8 years from now.
g) The value of $10,000 held in Account A for one year and then held in Account C for
one year.
146
→
W ORKSHEET #21
Interest
14 You deposit $2,500 in a bank. The annual interest rate is 8%. To compute the value of your
deposit after 212 days, several methods are available:
I. Daily compounding (assuming 360 days per year).
II. Compounding twice a day.
III. Continuous compounding.
IV. Compounding once a year.
The eight expressions below are candidates for evaluating the balance after 212 days by one
or the other method. Write next to each expression the number of the method the expression
represents. Not all expressions represent a method.
0.08
360
0.08
360
2500 1 +
2500 1 +
360/212
212
2500(1 + 0.08)212/360
2500 1 +
0.08
720
424/720
2500 · e(0.08)(212)/360
2500 1 +
0.08
720
424
2500 · e(1+0.08)
2500 1 +
0.08
212
212
360
Worksheet #22
Annual Percentage Yield
Key Question
Which of the following accounts is the most lucrative? (That is, which will yield the most interest?)
I. annual rate of 6%, compounded quarterly
II. annual rate of 5.9775%, compounded monthly
III. annual rate of 5.875%, compounded daily
IV. annual rate of 5.8125%, compounded continuously
From Worksheet #21, we observed that the more times the account is compounded, the more interest is
earned. Also, it seems obvious that a larger rate produces more interest. But how do we compare investments
that have different rates and different compounding periods?
In order to get a grasp of the strategy for comparing different investments, let’s take a side-trip down the
spaghetti aisle.
1
Information about two different brands of spaghetti sauce is listed in the table below.
Brand
Size (oz.)
Price ($)
Saucy McShea’s
28
2.39
Taste-Tastics
26
2.29
Saucy McShea’s is more expensive, but it’s also larger. For each brand, compute the price
per ounce to determine which of the sauces is a better buy. (Don’t worry about taste.)
Most markets will display the price per ounce on the shelves to provide shoppers with convenient information to compare brands. The key idea here is that the prices of different items need to be converted into a
common unit in order to compare them properly.
We need to do the same for our bank accounts. The common unit in this context is the Annual Percentage
Yield (APY). The APY is defined to be the percentage of the balance that the account earns in interest in
one year. The Truth in Lending Act requires all advertisements for a Certificate of Deposit to list its APY.
147
148
2
W ORKSHEET #22
Annual Percentage Yield
Consider the account in Key Question I: 6% interest, compounded quarterly.
(a) If you make an initial deposit of $500 into this account:
i. How much interest (in dollars) do you make in the first year?
ii. What percentage of your original deposit do you make in interest in the first year?
(b) If you make an initial deposit of $3112 into this account:
i. How much interest (in dollars) do you make in the first year?
ii. What percentage of your original deposit do you make in interest in the first year?
(c) If you make an initial deposit of $P into this account:
i. How much interest (in dollars) do you make in the first year?
ii. What percentage of your original deposit do you make in interest in the first year?
The punchline here is: no matter how much you deposit initially, you will earn 6.136% of your investment
in a year. This number is the APY for this account. (It’s important to note that an account advertised as 6%
annual interest, compounded quarterly, will actually gain more than 6% in a year.)
In general, the interest earned after one year on an account with annual interest rate r, compounded n
times a year will be:
I=
=
=
A(1) − P
r n×1
P 1+
−P
n
n
r
P 1+
−1
n
The APY is the percent of the principal that you get in interest:
APY =
=
=
I
× 100%
P
n
P 1 + nr − 1
× 100%
P
r n
− 1 × 100%
1+
n
This gives a nice formula for the APY. Notice that the formula does not depend on P .
3
Use the APY formula to compute the APY for the accounts in Key Questions II and III.
Things change a bit if interest is compounded continuously. Recall that the amount in the account after
t years is A(t) = P ert .
W ORKSHEET #22
4
Annual Percentage Yield
149
Suppose you deposit $P into an account paying (r × 100)% interest compounded continuously.
(a) How much interest (in dollars) do you earn in the first year?
(b) What percentage of your principal is this?
This gives a formula for the APY of an account paying (r × 100)% interest compounded continuously:
AP Y = [(er − 1) × 100%] .
Note that this formula depends only on the interest rate r.
5
Compute the APY for the account in Key Question IV.
6
Which of the accounts in the Key Question should you put your money in? Why?
Summary:
• The APY is the percentage of your balance that an account pays in interest in one year.
• Given a choice between several different accounts, choose the one with the highest APY.
• If the account bears (r × 100)% interest compounded n times a year, then the APY is given by the
formula:
r n
− 1 × 100%.
AP Y = 1 +
n
• If the account bears (r × 100)% interest compounded continuously, then the APY is given by the
formula:
AP Y = [(er − 1) × 100%] .
Worksheet #23
Solving for Time
Key Questions
I. If you deposit $10,000 into an account that pays 3.11% interest, compounded continuously,
how long until your balance is $17,000?
II. If you deposit $3,000 into an account that pays 1.74% interest, compounded quarterly, how
long until your balance is $6,000?
III. How long will it take to triple your principal in an account that pays 2.47%, compounded
monthly?
Let’s start with Key Question I. Using the formula for continuously compounded interest, we can set up a
formula for A(t), the balance in the account after t years:
A(t) = 10, 000e0.0311t .
We’re looking for the time at which A(t) reaches the value 17,000. In other words, we need to solve the
following equation for t:
17, 000 = 10, 000e0.0311t .
We can divide both sides of the equation by 10,000 to isolate the exponential part of the equation:
1.7 = e0.0311t .
But now the variable we need to solve for is in the exponent and we need a tool, the natural logarithm, to
solve for it.
The natural logarithm (abbreviated ln) “undoes” the action of the function ex .
calculator will tell you that
e2 = 7.389056099.
For example, your
The ln “undoes” this:
ln(7.389056099) = 2.
By itself, the statement “ln(7.389056099) = 2” tells you that 2 is the exponent to which you raise e in order
to get the number 7.389056099. In general,
if ex = y, then x = ln y.
1
Rewrite the following statements using the ln:
(a) e3 = 20.08553692
(b) e10 = 22, 026.46579
150
W ORKSHEET #23
Solving for Time
151
(c) e0.6931471806 = 2
(d) e2.302585093 = 10
2
Check that you understand how to use your calculator to compute a natural logarithm by
computing:
(a) ln(3)
(b) ln(400)
(c) ln(e6 )
The logarithm satisfies several nice properties:
(A) If a = b, then ln(a) = ln(b). (That is, you can take the ln of both sides of an equation.)
(B) ln(ab ) = b × [ln(a)]
(C) ln(e) = 1
(D) ln(a × b) = ln(a) + ln(b)
(E) ln
a
b
= ln(a) − ln(b)
(Properties (A), (B), (D), and (E) are true for all positve numbers a and b. Property (C) works only for that
special number e.)
If you do things right, you’ll never need to use properties (D) and (E) in this course. I include them here
because they are important properites if you’re ever working more generally with logarithms.
3
Back to Key Question I. When we left, we needed to solve the equation 1.7 = e0.0311t for t.
In the table below, state which of the properties of natural logarithms justifies each step.
Step
Property
1.7 = e0.0311t
N/A
ln(1.7) = ln(e0.0311t )
Property (A)
ln(1.7) = (0.0311t) × [ln(e)]
ln(1.7) = 0.0311t
Now use your calculator to compute the value of ln(1.7) and solve for t to answer Key
Question I.
4
Replicate the above process to answer the following question: If you deposit $400 into an
account paying 4.2% interest, compounded continuously, how long until your balance is
$500?
152
5
W ORKSHEET #23
Solving for Time
(a) In Key Quesiton II, give a formula in terms of t for A(t), the balance in the account
after t years.
(b) Set up the equation you need to solve in order to answer Key Question II.
(c) Divide both sides of your equation by 3000 (to isolate the exponential part) and then
use properties (A) and (B) from page 151 to solve for t, answering Key Question II.
Why don’t you need property (C)?
Answering Key Question III should be similar to Key Question II, but they don’t give us a value for P . We
have:
0.0247 12t
A(t) = P 1 +
OR A(t) = P (1.002058333)12t .
12
We want to know how long it will take for A(t) to be three times the principal. That is, when will A(t)
equal 3P ? We set the formula for A(t) equal to 3P :
3P = P (1.002058333)12t .
6
(a) Divide both sides of the above equation by P to isolate the exponential part and use
your logarithm properties to solve for t, answering Key Question III.
(b) Explain why they didn’t give you a specific principal in this Key Question.
→
7
(a) How long does it take to double the value of a dollar in each of the following accounts?
i.
ii.
iii.
iv.
11% per year, compounded quarterly
9% per year, compounded continuously
6% per year, compounded daily
8% per year, compounded monthly
(b) How long does it take for $1000 to be increased by 25% in an account that pays 8% per
year, compounded continuously?
(c) Mary and Charlie both have bank accounts. Mary’s earns 8% per year and Charlie’s
earns 7% per year (both compounded annually). Mary puts A dollars into her account.
On the same day, Charlie deposits 15% more than Mary. At what time, if ever, will the
two accounts have the same balance?
→
8
Money is held in two accounts. The descriptions of the accounts are as follows:
Account A: Compounded continuously at an annual rate of (r · 100)%
Account B: Compounded monthly at an annual rate of (r · 100)%.
Answer each of the following questions for both accounts.
W ORKSHEET #23
Solving for Time
153
a) By what factor is the balance at the end of 2 years multiplied to get the balance at the
end of 2.5 years?
b) Suppose that the balance is doubled every 7 years. What is the value of r?
c) Find the value of r that makes the following sentence true: The annual percentage yield
(APY) of this account is (s · 100)%.
d) Suppose r = 0.05 and you have $10,000 to deposit. Compute how much interest the
account will earn in 2 years and when the account will have earned exactly half that
much interest.
→
9
A bacteria colony doubles its population every 20 minutes. At the start of an experiment,
there are P million bacteria in the colony.
(a) In terms of P , how many bacteria are in the colony after 20 minutes? after 40 minutes?
after 60 minutes?
(b) In terms of P , how many bacteria are in the colony after one hour? after 2 hours? after
3 hours? after t hours?
(c) How long does it take to triple the population?
154
W ORKSHEET #23
Solving for Time
Index
additive sequence, 120
annual percentage yield (APY), 147
average cost (AC), 58, 60
average revenue, 33
average speed, 15
average trip speed, 15
average variable cost (AVC), 59, 60
breakeven price, 57, 58
breaking even, 69
CGPA, 34
compound amount formula (CAF), 129
compound interest, 128
compounding periods, 128
continuous compounding, 144
cost
average (AC), 58, 60
average variable (AVC), 59, 60
marginal (MC), 31, 62
total (TC), 29, 58
variable (VC), 58
delta (∆) notation, 17
diagonal line, 18
e (2.71828...), 144
explicit formula, 121
fixed cost (FC), 57
functional notation, 39
functions
quadratic, 89
future value, 132
GPA, 34
incremental speed, 33
instantaneous rate of change, 15
interest
annual percentage yield, 147
compound, 128
compound amount formula, 129
continuously compounded, 144
partial compounding periods, 135
line
diagonal, 18
parallel, 51
secant, 18
ln, 150
logarithm, 150
marginal analysis, 80
marginal cost (MC), 31, 62
marginal revenue (MR), 30, 62
market price, 56
maximizing profit, 61, 64
maximum possible profit, 31
MC (marginal cost), 31
MR (marginal revenue), 30
multiplicative sequence, 122
natural logarithm (ln), 150
notation
functional, 39
overall speed, 33
parallel lines, 51
partial compounding periods, 135
present value, 132
price
shutdown, 58
155
156
principal, 128
profit, 29
how to maximize, 61, 64
maximum possible, 31
proportionate change, 123
puppies, 52
QGPA, 34
quadratic formula, 8
quadratic functions, 89
rate of change, 15
incremental, 15
instantaneous, 15
overall, 15
recursive formula, 121
revenue
average, 33
marginal (MR), 62
total (TR), 28
reveune
marginal (MR), 30
rolling ruler method, 19
secant line, 18
sequence, 119
additive, 120
explicit formula, 121
multiplicative, 122
recursive formula, 121
shutdown price, 58
slope, 18
speed, 15
average, 15
formula for, 15
incremental, 33
overall, 33
TC (total cost), 29
total cost (TC), 29, 58
total revenue (TR), 28
TR (total revenue), 28
trip speed, 15
average, 15
I NDEX
variable cost (VC), 58
vertex formula, 8
Appendices
157
G LOSSARY
159
Glossary
Annual Percentage Yield (APY) The percentage change in the balance of an account over a one
year period.
. Graphically, average
Average Cost (AC) A company’s cost per quantity sold: AC(q) = TC(q)
q
cost at q is the slope of a diagonal line through the total cost graph at q.
Average Revenue (AR) A company’s revenue per quantity sold: AR(q) = TR(q)
. Graphically,
q
average revenue at q is the slope of a diagonal line through the total revenue graph at q.
Average Speed Distance traveled divided by time elapsed. Given by the slope of a secant line
through the distance graph.
Average Trip Speed Total distance travelled divided by time elapsed. If the distance graph goes
through the origin, then average trip speed at time t is given by the slope of the diagonal line
through the distance graph at t.
=
Average Variable Cost (AVC) A company’s variable cost per quantity sold: AVC(q) = VC(q)
q
TC(q)−FC(q)
.
q
Graphically, average variable cost at q is the slope of a diagonal line through the
variable cost graph at q.
Breakeven Price The transitional market price for a company to make a profit: if the market price
p is lower than the breakeven price, the company can’t make a profit. Graphically the breakeven
price is the slope of the least steep diagonal line that intersects the total cost graph, i.e. the slope
of the diagonal line which is also a tangent line to the total cost graph. The breakeven price is the
smallest value of AC. It is also the price where the marginal cost graph and the average cost graph
intersect. (Note: It is important to remember that the breakeven price is a price. )
Compound Interest Interest paid on interest reinvested: If at the end of a payment period the
interest due is reinvested at the same rate, then the interest as well as the original principal will
earn interest during the next payment period.
Demand Curve/Demand Function Shows the relationship between the quantity of a product demanded, q, and its price, p. The demand function gives price as a function of quantity. It frequently
is a straight line p = f (q) = mq + b, where the slope is negative, since the law of demand states:
“Other things remaining the same, the higher the price of a good, the smaller is the quantity demanded.” There is a subtlety here, the quantity demanded depends on the price, however, contrary
to mathematical convention, the demand curve is made with quantity on the horizontal axis, and
the demand function is given as a function of quantity.
Diagonal Line A line that goes through the origin.
Fixed Cost (FC) The cost of all items which do not depend on the quantity produced, i.e. the
value of total cost when q = 0. Fixed cost includes costs like building and equipment rental, loan
payments, some salaries, etc.
Future Value The value of an investment after a given time. When using the CAF, the future value
is the amount A.
Marginal Analysis A technique for finding maximum profit by comparing marginal revenue with
marginal cost. If MR > MC, then the extra revenue from selling one more unit exceeds the extra
cost incurred to produce it, so the profit increases if output increases. If MR < MC, then the
extra revenue from selling one more unit is less than the extra cost, so the profit decreases if output
160
A PPENDICES
increases. If MR = MC then profit is maximized. Given the graphs of total revenue and total cost,
one can find maximum profit by comparing the secant lines (or tangent lines). Maximum profit
occurs when the tangent line of the total revenue graph at q is parallel to the tangent line of the
total cost graph at q.
Marginal Cost (MC) The change in total cost that comes with an increase of one unit is sales:
MC(q) = TC(q + 1) − TC(q). Graphically, marginal cost is the slope of a secant line to the total
cost graph, where the q values are one apart. Thus, marginal cost can be estimated by finding the
slope of the tangent line to the total cost graph at q.
Marginal Revenue (MR) The change in total revenue that comes with an increase of one unit is
sales: MR(q) = TR(q + 1) − TR(q). Graphically, marginal revenue is the slope of a secant line to
the total revenue graph, where the q values are one apart. Thus, marginal revenue can be estimated
by finding the slope of the tangent line to the total revenue graph at q.
Percent Change If a quantity changes from an old value O to a new value N , then percent change
N −O
is:
× 100%.
O
Present Value The value of an investment today. When using the CAF, the present value is the
initial principal.
Profit The difference in total revenue and total cost for a company: P= TR − TC.
Proportionate Change Percent change, expressed as a decimal.
Secant Line A line that goes through two points on a graph.
Shutdown Price A transitional market price: if the market price p is less than the shutdown price,
the company will always lose more than their fixed cost regardless of the quantity sold; if p is
greater than the shutdown price, then there are some quantities that will allow the company to
make back some (if not all) of their fixed costs. Graphically the shutdown price is the slope of the
least steep diagonal line that intersects the variable cost graph, i.e. the slope of the diagonal line
which is also a tangent line to the variable cost graph. The shutdown price is the smallest value
of AVC. It is also the price where the marginal cost graph and the average variable cost graph
intersect. (Note: It is important to remember that the shutdown price is a price. )
Total Cost (TC) The total amount of money a company spends: TC = FC + VC, where FC is
fixed cost and VC is variable cost.
Total Revenue (TR) The total amount of money that a company brings in. Total revenue is equal
to the price of the output multiplied by the number of units of output sold, that is TR = pq, where
p is price and q is quantity.
Variable Cost (VC) The amount it costs for a company to produce q items without including the
fixed cost, that is, costs incurred by the company for the variable inputs it uses. Examples of
variable cost are raw material costs and some labor costs. Graphically, the variable cost graph
always goes through the origin and has the same shape as the total cost graph.
A NSWERS TO S ELECTED E XERCISES
161
Answers to Selected Exercises
These answers are intended for you to use mostly to check your work. You should be careful
to work each exercise before checking your answers. Make corrections, if necessary, and write
up complete solutions to each exercise to turn in. If you don’t understand how to get the correct
answer, ask your TA, your instructor, or a tutor in the Math Study Center.
PROLOGUE
1 A) -2.4; B) 13 ; C) 21 ;
3 yes
4 yes
6 A: f (x) = 2x − 2; B: f (x) = 21 x + 1; C: f (x) = − 32 x + 11; D: f (x) = − 52 x + 6;
7 a) s = − 31 , t =
16
;
9
b) p = 3, q = 0; c) x = 83 , y = 76 ;
8 a) x = 2 and x = −5; b) x = 3 and x = 4; c) q =
√
√
3
2
and q = −4; d) no solutions;
9 a) r = 6±4 68 = 3±2 17 ; b) no
solutions; c) x = 2 and x = −5; d) x =
√
√
−6± 60
e) no solutions; f) t =
= −3 ± 15
2
√
6± 28
2
=3±
√
7;
WORKSHEET #1
2 (a) 24 miles; (b) approximately 21.3 miles;
4 (d) is true;
7 from t = 30 to t = 60 minutes;
8 Average trip speed goes down while approximate speed goes up from approximately t = 30
to t = 45 minutes.
13 (a) approximately 4 miles;
(b) approximately 0.45 mi/min;
(c) approximately 0.1 mi/min;
(d) any 5-minute interval starting later than 45 minutes;
(e) t = 40;
(f) ave speed=approximately 0.3 mi/min, distance traveled=approximately 0.6 miles;
(g) 15 to 20 minutes or 40 to 45 minutes;
(h) approximately 0.32 mi/min (occurs at t = 45 minutes).
162
A PPENDICES
14 (a) 45 miles;
(b) At t = 180, car A was behind car B. At t = 240, car A had caught up to car B. Thus, car
A traveled further than car B during that hour.
(c) t = 230 mins.;
(d) 60.5 miles;
(e) False;
15 (a) approximately 12 mins.;
(b) 30 miles;
(c) 42.5 miles;
(d) approximately 1.8 mpm;
(e) B, because the average trip speed (slope of the diagonal to the distance graph) is lowest
around t = 22.
WORKSHEET #2
1 Between 3:00 and 3:30, 630 gallons were drawn from the reservoir.
2 6100 gallons; 3760 gallons;
6 at 1:00; 2.1 thousand gallons;
7 (a) 6.16 thousand gallons; (b) 5.16 thousand gallons;
8 (a) from 0 to 1 and from 2 to 3;
(b) Fourth graph from left;
(c) (i) t = 0.5, (ii) t = 2, (iii) t = 4;
(d) from t = 2.5 to t = 3;
(e) −0.9 inches;
(f) 0.5 inches.
9 (a) 1.2 inches;
(b) 64.8 inches;
(c) from 12 to 13;
(d) (ii);
(e) He was 62.5 inches six months after 13th birthday. So lifetime growth rate = 0.39
inches/mo.
10 (a) t = 8.8 or 20.4; (b) 197.17 gallons per hour;
(c) 20.83 gallons per hour; (d) 1300 gallons;
A NSWERS TO S ELECTED E XERCISES
163
WORKSHEET #3
1 $15; $28; $39;
2 T R = p × q;
4 q = 8 reams; $64; no;
5 T C = 4q;
7 If you sell more than 12 reams, you lose money.
9 q = 6 reams; P = $36;
WORKSHEET #4
3 M R at q = 4;
9 QGP = 56; QCr = 17; QGP A = 3.3;
12 Compute the slope of the diagonal line through the CGP curve at Cr = 60. This line is both
a diagonal line and a secant line.
13 Compute the slope of the secant line through the CGP curve at Cr = 45 and Cr = 60. This
is a secant line but not a diagonal line.
15 (a) approximately 321 cars/hr;
(b) from t = 7 to t = 8;
(c) 2,000 cars at 5 hrs., 250 cars at 10 hrs.;
(d) approximately 333 cars/hr.;
(e) most at t = 4 (approximately 2050 cars), fewest at t = 10 (approximately 250 cars).
16 (a) $99.16; (b) $18.61; (c) from q = 0 to q = 8
WORKSHEET #5
1 Purple travels 0 miles in the first 5 minutes. Purple reaches 8.5 miles at t = 10 minutes.
Purple reaches 14.6 miles at t = 15 minutes.
2
(a) Purple’s distances would be Red’s distances shifted right three boxes.
(b) Purple’s distances would be Red’s distances shifted left one box.
3
(b) P is R shifted right 5 units.
(c) P would be R shifted right 15 units.
(d) P would be R shifted left 5 units.
164
A PPENDICES
5 t = 32.5;
7
(a) The red car travels 18.7 miles in the first 15 minutes.
(b) R(25) = 23.0;
8 (a) Total revenue for 2 reams is $28. (b) f (4) = 48;
9
(a) The red car travels 2.7 miles from t = 30 to t = 45 minutes. (False)
(b) The red car has traveled further after 30 minutes than after 25 minutes. (True)
10
(a) R(40) − R(25) = 3.7 (True)
(b) R(20) > R(15) (True)
12 D(20) = R(20) − R(15)
14 (b) 1,000 cars; (d) approximately 2,600 cars.
WORKSHEET #6
1 (a) 22 thousand gallons; (b) 15.5 thousand gallons; (c) 2.8 thousand gallons per hour;
(d) 2 thousand gallons per hour;
2 (i)=(c); (ii)=(a); (iii)=(b); (iv)=(d);
3
(a) 24.8 thousand gallons — the amount that has flowed in from 3 hours to 10 hours;
(b) 1.5 thousand gallons per hour — the overall average rate of flow at t = 5 hours;
(c) 5.5 thousand gallons per hour — the incremental average rate of flow over the two-hour
interval starting at t = 6;
7
English
Functional Notation
Graph Language
the amount that
flows in from t =
1 to t = 4
A(4) − A(1)
the change in
height from t = 1
to t = 4
the overall average rate of flow
after 5 hours
A(5)
5
the slope of the
diagonal line at
t=5
Continued on next page
A NSWERS TO S ELECTED E XERCISES
165
continued from previous page
8
English
Functional Notation
Graph Language
the incremental
average rate of
flow from t = 1
to a time h hours
later
A(1+h)−A(1)
h
the slope of the
secant line from
t = 1 to t = 1 + h
the amount that
flows in from t =
1 to t = T
A(T ) − A(1)
the change in
height from t = 1
to t = T
the overall average rate of flow
after 4 hours
A(4)
4
the slope of the
diagonal line at
t=4
the incremental
rate of flow from
t = 3 to a time h
hours later
A(3+h)−A(3)
h
the slope of the
secant line from
t = 3 to t = 3 + h
(a) (i) 0; (ii) 0; (iii) 9;
(b) (i) 6 to 8; (ii) 7 to 9; (iii) 2 to 4;
(c) (i) Any x greater than 1.5; (ii) 0 to 3; (iii) 13.5;
(d) (i) B; (ii) D; (iii) C.
WORKSHEET #7
1 t = 6.75
2 4 hours
3 t = 5.75
5 t = 4.5 or t = 7.5
6 Functional notation: Find a value of t such that A(t + 1.5) − A(t) = 6.
7 from 4.5 to 6 or from 6.8 to 8.3
8 (a) 0.6 lbs.;
(b) t = 48 to t = 54;
(c) t = 50;
166
A PPENDICES
(d) approximately 0.05 lbs/hr;
(e) 1.7 lbs. at 66 hrs.;
(f) 0.53 lbs.
9 (a) Straight line, starts at 10 and has slope 5;
(b) TG inc by $10, ST inc by $10;
(c) 1.5 months or 9.5 months;
(d) $1.43/mo.;
(e) 2 to 4 months;
(f) 21 to 2 12 months, or 10 to 12 months;
(g) 10 months;
(h) ST.
10 (a) straight line starting at 0 with slope 2;
(b) approximately 4,000 gals;
(c) approximately t = 7.75;
(d) About 6:30 AM;
(e) t = 3 and t = 8;
(f) t = 6 to t = 8.5;
(g) Anytime after t = 8.
11 (a) 0.8 or 6.3;
(b) 7;
(c) 1.1 or 8.
WORKSHEET #8
1
(a) T R(100) = $75; T R(300) = $225; T R(600) = $450;
(b) slope=0.75; intersects the vertical axis at the origin;
(c) q = 270 to q = 700;
2
(a) T R(100) = $35; T R(300) = $105; T R(600) = $210;
(b) slope=0.35; intersects the vertical axis at the origin;
(c) No production level will yield a profit.
3 yes; between q = 350 and q = 660;
5 $0.49;
6
I. any market price greater than $0.49;
II. any market price less than (or equal to) $0.49;
7
(a) $115; this is more than the F C;
A NSWERS TO S ELECTED E XERCISES
167
(b) $70; this is less than the F C;
(c) between q = 370 and q = 530;
8 no;
10 $0.32;
11 AC(300) = $0.67 per bag;
12 AC decreases then starts to increase.
15 (a) AVC = $6.00/Blivet, AC = $10.00/Blivet;
(b) MR=$11;
(c) 45;
(d) $180;
(e) None (MR is always decreasing);
(f) q = 78;
(g) q = 92
WORKSHEET #9
1 (b) q ≈ 550; (c) $135;
2 (a) q ≈ 550; (b) $215;
(c) The vertical distance between T C and V C is always the same. The vertical distance
between T R and V C is always 80 more than the vertical distance between T R and T C. If
the vertical distance between T R and T C gets larger or smaller, so does the vertical distance
between T R and V C.
4
(a) M R(100) = $1; M R(300) = $1; M R(600) = $1;
(b) The secant line through any two points on the T R graph is the T R graph itself.
5 M C(100) = $0.45; M C(300) = $0.08; M C(600) = $1.00;
6 q ≈ 600; no—the T R graph is off the chart;
7
(a) M R(100) = $0.80; M R(300) = $0.80; M R(600) = $0.80;
(b) Sketch a horizontal line 0.80 units above the horizontal axis.
(c) q ≈ 560;
8 q ≈ 580;
10 The BEP is both the slope of a diagonal line (a value of AC) and the slope of a tangent line
(a value of M C).
168
A PPENDICES
11 The graphs of T C and V C have exactly the same shape. All their secant lines and tangent
lines are parallel.
13 (a) $510;
(b) $0.75;
(c) $1.25;
(d) $2.08;
(e) q = 510.
14 (a) $1.49;
(b) $2.10;
(c) $400;
(d) $155.
WORKSHEET #10
1 The selling price and marginal cost are both constants. Each time we sell one more machine,
our total revenue goes up by $40 and our total cost goes up by $15.
2 (b) q = 2 thousand machines;
3 q = 5 thousand machines;
4 q = 1.4 thousand machines;
5 q = 3 thousand machines;
6 (a) T RII (q) = 40q; T CII (q) = 30q + 50;
7 T RIII (q) = 40q; T CIII (q) = 15q + 35; T RIV (q) = 24q; T CIV (q) = 10q + 42;
8 (a) T RV (q) = pq; T CV (q) = mq + c; (b) q =
9 (b)
c
;
p−m
(i) q gets larger
(ii) q gets smaller
(iii) q gets smaller
(iv) q gets smaller
(v) q gets larger
(vi) q gets larger
10 Raise p to $55 per machine.
WORKSHEET #11
1 These are approximated from the graph:
(a) 9.6 miles; (b) 7; (c) 0.66 mpm; (d) 0.75; (e) 0.58; (f) 0.15;
2 from graph: D(11) ≈ 8; from formula: D(11) = 7.975;
3 D(16) = 9.6; D(9) = 6.975;
A NSWERS TO S ELECTED E XERCISES
4
D(14)
14
= 0.65;
D(10)
10
169
= 0.75;
5 HINT: It may help to draw a reference line with slope 0.7 =
7
10
=
rise
.
run
4
10
=
rise
.
run
7 t = 12;
9 HINT: It may help to draw a reference line with slope 0.4 =
11 (a) D(t + 5) − D(t) = −0.25t + 4.375 = 4.375 − 0.25t; (b)
12
D(6+5)−D(6)
5
= 0.575;
D(15+5)−D(15)
5
D(t+5)−D(t)
5
= 0.875 − 0.05t;
= 0.125;
13 t = 9.5;
14 (a)
15
D(t)
t
= 2 − 0.04t; (b) t = 17; (c)
D(t+2)−D(t)
2
= 1.92 − 0.08t; (d) t = 11.
(a) 4a + 2b + 8 = 8 and 16a + 4b + 8 = 0, so a = −1 and b = 2;
(b) f (x) = −x2 + 2x + 8, f (3) = 5, f (0) = 8, so
(c)
f (x)−f (0)
x
=
−x2 +2x+8−8
x
f (3)
3
=
5
3
and
f (3)−f (0)
3
= −1;
= −x + 2;
(d) f (3.5) − f (3) = (−(3.52 ) + 2(3.5) + 8) − (−(32 ) + 2(3) + 8) = 2.75 − 5 = −2.25;
(e) f (x + 2) = (−1)x2 + (−2)x + 8;
16
(f)
f (x+2)−f (x)
2
(a)
I(t)
t
= −2x.
= at + b;
(b) 10 = 10a + b and 10 − 10a + b, so a = −1 and b = 20;
(c) I(t) = −t2 + 20t, I(4) − I(3) = 13 gallons;
(d) Amount flowed out at time t is O(t) = 8t. So Amount in reservoir is I(t)−O(t)+10 =
−t2 + 12t + 10.
17
(a)
(b)
(c)
A(t)
= − 12 t + 15; B(t)
= 5;
t
t
− 12 t + 15 = 5 when t = 20
− 21 t + 15 = 12 when t = 6
minutes;
minutes;
(d) B(t) = 5t = 40 when t = 8 and
A(8)
8
= 11 feet per minute.
WORKSHEET #12
1 p = −q + 16 OR p = 16 − q;
2 T R(q) = −q 2 + 16q OR T R(q) = 16q − q 2 ;
5 q = 5.5;
170
A PPENDICES
6 M R = 6 at q = 4.5. So, profit is maximized at q = 5 reams.
7 p = 17 − q;
8 T R(q) = 17q − q 2 ;
9 M R(q) = 16 − 2q;
10 q = 6 (q = 7 is also an acceptable answer — why?);
12 p = 12 − 0.5q; T R(q) = 12q − 0.5q 2 ; M R(q) = 11.5 − q; M C(q) = 3.25; M R = 3.25 at
q = 8.25; Profit is maximized at q = 9 reams.
13
(a) M R(q) = − 12 q + 29.75 and M C(q) = 17.5;
(b) − 21 q + 29.75 = 17.5 when q = 24.5. So, q = 25 Trivets maximizes profit.
(c) M R(q) = M C(q) + 5.25 ⇒ − 12 q + 29.75 = 17.5 + 5.25 ⇒ q = 14;
14
(d) AR =
TR
q
(e) AC =
TC
q
= − 41 q + 30; AR = 16 when q = 56;
= 17.5 +
100
;
q
AC = 21.50 when q = 25.
√
√
(a) MR(m) = T R(m + 0.001) − T R(m) = 5 m + 0.001 − 5 m;
√
(b) q + 1 = 5 q;
(c) AC =
TC
= 1 + 1q = 1.25 when q = 4 thousand Blivets;
q
√
5 q
= 1.30 when q = 14.8 thousand Blivets.
q
(d) AR =
√
(e) 5 q − q − 1 = 4.
WORKSHEET #13
2 t = 5 hours;
3 12.5 gallons;
4 D(t) = − 23 t2 + 11t − 9; since the coefficient in front of the t2 is negative, the vertex gives a
high point; t = 11
;
3
5 D
6
3 2
t
2
11
3
=
67
6
= 11.17 gallons;
− 11t + 9 = 0 OR − 23 t2 + 11t − 9 = 0; t = 0.94 hours;
9 R5 (t) = 21 t2 − 5t + 30; R15 (t) = 21 t2 − 5t + 40; R−10 (t) = 12 t2 − 5t + 15; The graphs of
R(t), R5 (t), and R−10 (t) have the same shape. R5 (t) is the graph of R(t) shifted up 5 units.
R−10 (t) is the graph of R(t) shifted down 10 units.
A NSWERS TO S ELECTED E XERCISES
171
10 Dd (t) = G(t) − Rd (t) = (−t2 + 6t + 16) − ( 21 t2 − 5t + 25 + d) = − 23 t2 + 11t − 9 − d;
11 t = 5 hours; Rd (5) = 12.5 + d gallons;
12 t =
11
3
hours; Dd
11
3
= 11.17 − d gallons;
√
11± 121−6(9+d)
(Depending on the size of the shift, this could give 0, 1, or 2 solutions.
13 t =
3
Key Question V asks for the smallest of these solutions.)
17 Rh (t) = 21 (t − h)2 − 5(t + h) + 25;
18 a = 21 , b = −(5 + h); c = 12 h2 + 5h + 25;
19 t = 5 + h hours;
21 R1 (t) = 21 t2 − 6t + 30.5; D1 (t) = (−t2 + 6t + 16) − ( 12 t2 − 6t + 30.5) = − 23 t2 + 12t − 14.5;
KQ III: t = 4 hours;
KQ IV: D1 (4) = 9.5 gallons;
KQ V: t = 1.48 hours.
22 (a)1.80 minutes;
(b) 1.93 minutes;
(c) G(t) = 1.3194t + 5;
(d) 18.2 miles;
(e) 0.28 miles;
(f) 2.12 minutes;
(g) (i) H(t) = 3t − 18 t2 + 1, (ii) 1 mile.
23 (a) at t = 4; 4 gallons;
(b) 1.55 and 6.45;
(c) 12 gallons;
(d) Vat B would never have as much water as Vat A.
(e) t = 4.25.
(f) any time between 4 and 4.25.
24 (a) 4.5;
(b) x = 0.90 and 4.43;
;
(c) x = 2 and x = 10
3
(d) from x = 1 to x = 38 ;
(e) from x = 2 to x = 4;
(f) from x = 1.55 to x = 6.45.
25 (a) 20 feet;
(b) t = 2.40;
172
A PPENDICES
(c) t = 0.9375;
(d) Greatest height reached is 34.0625 ft. It hits the ground after about 2.4 seconds;
(e) t = 0.43 seconds;
(f) t = 1.875. [Note: This is twice the time it took the rock to reach it’s highest point. Does
that make sense?]
26 (a) D(t) = 3 + 6t;
(b) 24 ft;
(c) t = 2 sec.
(d) (i) 2t + 1 = 3t + 6; (ii) 2t2 − 5t − 3 = 0.
(e) First car is 5 feet ahead at t = 12 and t = 2 seconds. Other car is 5 feet ahead at t = 3.61.
27 (a) q(3) = 5;
(b) q(x) = −x + 4 + 12
;
x
(c) The correct graph is the one on the right.
(d) x = 3;
(e) 15.
(f) (i) −x + 4 + 12
= 2x + 4. (ii) 3x2 − 12 = 0; (iii) x = 2.
x
28 (a) 14.96 gallons;
(b) 4.90 gal/hr.
(c) t = 10.06, t = 0.0062;
(d) 16t2 − 89t + 100 = 0;
(e) 34 gallons.
WORKSHEET #14
1 S(2) = 14; S(4) = 24; S(6) = 30; S(8) = 32;
3 C(2) = 10; C(4) = 14; C(6) = 18; C(8) = 22;
6 D(x) peaks first;
7 S(x) = − 21 x2 + 8x; C(x) = 6 + 2x;
8 D(x) = − 12 x2 + 6x − 6;
9 Both open down. S(x) hits its peak at x = 8. D(x) hits its peak at x = 6.
11 T R(q) = −0.6q 2 + 10q; T R is largest at q = 8.33.
12 T C(4) = 24; T C(6) = 32; T C(8) = 40;
13 T C(q) = 8 + 4q;
A NSWERS TO S ELECTED E XERCISES
173
14 Profit = T R(q) − T C(q) = −0.6q 2 + 6q − 8; Profit is largest at q = 5.
16 Profit is maximized at q = 6 reams. Maximum profit is $15.20.
17 (a) A(x) = 12 + 6x, B(x) = 2x2 + x;
(b) 14;
(c) C(x) = −2x2 + 5x + 12, C(2) = 14;
(d) x = 1.25;
(e) from x = 1.25 to 4;
(f) x = 2;
(g) x = 4.
√
18 (a) A(r) = 5 r + 2r;
(b) 4r2 − 49r + 36 = 0;
(c) 9.375;
q
(d) r = 3 25
≈ 1.16.
16
WORKSHEET #15
1
(b)
(c)
4
1 3
3 2
q − 10
q + q + 0.8;
30
3
1 2
q − 10
q + 1;
AV C(q) = 30
1 2
3
AC(q) = 30
q − 10
q + 1 + 0.8
;
q
(a) T C(q) =
I.
II.
III.
IV.
1 3
q
30
1 2
q
30
1 3
q
30
1 2
q
10
−
−
−
−
3 2
q + q + 0.8 = 3.20;
10
3
q + 1 = 0.55;
10
3 2
q + q = 0.55q;
10
3
q + 1 = 0.65;
5
5 190 framits (q = 1.90);
6 710 framits (q = 7.10);
8 66 framits (q = 0.66);
9
V. $0.10;
VI. q = 5.55;
VII. $0.325;
1 2
3
1 2
VIII. BEP occurs when 30
q − 10
q + 1 + 0.8
= 10
q − 35 q + 1. Multiplying both sides by q
q
3 2
1 3
1 3
q − 10
q + q + 0.8 = 10
q − 35 q 2 + q. This is a cubic equation we can’t solve.
gives: 30
From the graph, it looks as though the BEP is $0.50.
174
A PPENDICES
10 (a) q = 6.944 thousand framits;
(b) T R(q) = 81
q.
7
4
(c) q = 7 ;
(d) Profit is largest at q = 53
. The largest profit will be 10.332 thousand dollars;
14
(e) q = 1.704 or 5.867;
(f) New T C(q) = q 2 + 4q + 12;
53
;
(g) q = 14
(h) If FC was greater than $14,332, then you would not make a profit at any value of q.
11 (a) p = 24.5 − 0.5q;
(b) T R(q) = 24.5q − 0.5q 2 ;
(c) $14;
(d) T C(q) = 100 + 5q;
(e) max profit occurs at q = 19 or q = 20;
(f) $90.
WORKSHEET #16
1 (a) A, C, B; (b) A, B, C; (c) B, A, C;
2 at t = 4: A travels at 2 feet per minute, B travels at 2.4 feet per minute, C travels at 4 feet
per minute;
at t = 5: A travels at 2.5 feet per minute, B travels at 2 feet per minute; C travels at 4.5 feet
per minute;
3 Car B is further ahead since they were next to each other at t = 0 and, from t = 0 to t = 4.5,
Car B goes faster than Car A.
4
I. At t = 0, the two cars are next to each other. From t = 0 to t = 4.5, B goes faster
than A, so B is getting further ahead. At t = 4.5, Car A begins to go faster than B and
starts to catch up to B. The cars are furthest apart at t = 4.5.
II. At t = 0, the two cars are next to each other. Car C is always going faster than Car A,
so Car C gets further and further ahead of Car A as the 10 minutes go by. The cars are
furthest apart at t = 10.
5 Car B:
4+3.6
2
= 3.8 feet per minute; Car C:
2+2.5
2
= 2.25 feet per minute;
6 Car B: 3.8 feet; Car C: 2.25 feet;
7
A: average speed =
min = 1 foot;
0+1
2
= 0.5 feet per minute; distance traveled = 0.5 feet per minute× 2
= 3.6 feet per minute; distance traveled = 3.6 feet per minute×
B: average speed = 4+3.2
2
2 min = 7.2 feet;
A NSWERS TO S ELECTED E XERCISES
C: average speed =
min = 5 feet;
8
(a) ave speed =
= 9 feet;
0+3
2
2+3
2
175
= 2.5 feet per minute; distance traveled = 2.5 feet per minute× 2
= 1.5 feet per minute; distance traveled = 1.5 feet per minute × 6 min
(b) ave speed = 1.6+0
= 0.8 feet per minute; distance traveled = 0.8 feet per minute × 4
2
min = 3.2 feet;
= 4.75 feet per minute; distance traveled = 4.75 feet per minute × 5
(c) ave speed = 3.5+6
2
min = 23.75 feet;
9
B: ave speed =
C: ave speed =
4+(4−0.4)
= 8−0.4t
= 4 − 0.2t; DB (t) = 4t − 0.2t2 ;
2
2
2+(2+0.5t)
= 4+0.5t
= 2 + 0.25t; DC (t) = 2t + 0.25t2 ;
2
2
12 C catches up to B at t = 4.44 minutes. So, any time interval that contains t = 4.44 is
correct, like from t = 4 to t = 5.
13
(a) S(0) = b, S(t) = b + mt;
b+(b+mt)
2
m 2
+ 2t ;
(b) ave speed =
(c) D(t) = bt
=
2b+mt
2
=b+
m
t;
2
14 DJ (t) = t + 0.4t2 ; DK (t) = 5t − 0.3t2 ; DJ (7) = 26.6; DK (7) = 20.3;
Car J is ahead by 6.3 feet.
15 (a) S(t) = −0.5t + 2.5;
(b) D(t) = −0.25t2 + 2.5t;
(c) 4 miles;
(d) Av Tr Sp = −0.25t + 2.5;
(e) D(r + 1) − D(r) = −0.5r + 2.25
(f) DYellow (t) = −0.25t2 + 2.5t + 2.
16 (a) a = −1, b = 18, c = 0.5, d = 4.5;
(b) $67.50;
(c) Profit increases by $7.50;
(d) AR(q) = −0.5q + 18.5; AC(q) = 0.25q + 4.25;
(e) New T R is T R(q) = −0.5q 2 + 17.5q.
17 (a) A(t) = −0.75t2 + 12t and B(t) = 0.375t2 + 3t;
(b) farthest at t = 4, 18 yds.;
(c) 17.1111 yds;
(d) 9.75 yds/sec;
(e) t = 8 sec;
(f) 4.3125 yds;
(g) B(r + 1) − B(r) = 0.75r + 3.375.
176
A PPENDICES
18 (a) M R(q) = 11 − 0.5q; T R(q) = −0.25q 2 + 11.25q;
(b) $33;
(c) Profit = P (q) = −0.25q 2 + 8.25q − 20.
(d) q = 22.5;
(e) q = 16.5;
(f) The interval from q = 16.5 to q = 22.5;
+ 3;
(g) Av Cost= 20
q
(h) q = 17.
WORKSHEET #17
1 A: 90, 100; B: 99, 106; C: 486, 1458; D: 6.25, 3.125;
2 B is additive. Its increment is 7. C is not additive because 2 + 4 = 6, but 6 + 12 = 18 —
you don’t get from one term to the next by adding the same number. (What about D?)
3 (d) is the only incorrect statement.
4 (a) 7; (b) 7; (c) 7; (d) 7; (e) 7; (f) 7; (g) 7; (h) 14; (i) 21;
5 A(100) = 1040; A(1003) = 10, 070; A(1, 000, 007) = 10, 000, 110;
6 B(k) = 64 + 7k; B(100) = 764; B(1003) = 7085; B(1, 000, 007) = 7, 000, 113;
7 (a) 3; (b) 3; (c) 3; (d) 3; (e) 3; (f) 32 ; (g) 33 ; (h) 33 ;
8 The multiplier of sequence C is 3. Its recursive formula is C(k + 1) = 3 × C(k). The
multiplier of sequence D is 0.5 or 21 . (What is its recursive formula?)
9 C(10) = 118, 098; C(19) = 2, 324, 522, 934 = 2.325 × 109 ; C(99) = 3.436 × 1047
10 D(k) = 100×(0.5)k ; D(10) = 0.09765625; D(19) = 1.907349×10−4 ; D(99) = 1.57772×
10−28 ;
12 Aginit is the best investment after 6 periods and after 12 periods.
13 At the end of year 4, your salary should be $43,297.29. The dollar amount of your raise that
year would be $865.95.
14 multiplicative; multiplier=1.02;
15 (a) S(k + 1) = 1.02 × S(k); (b) S(k) = 40, 000 × (1.02)k ; S(10) = $48, 759.78;
(c) 0.2190; (d) 21.90%;
16
(a) (i) A(k + 1) = A(k) + 5; (ii) B is additive with increment 10; (iii) B(108) = 1092;
(b) (i) C(k + 1) = 9C(k); (ii) D is multiplicative with multiplier 3; (iii) D(k) = 110 · 3k .
A NSWERS TO S ELECTED E XERCISES
17 (a) n = m3 or m = n1/3 ;
(b) m6 ;
177
(c) n2/3 ;
(d) mj ;
(e) = , < , = .
18 (a) 3.48;
(b) f (3) = 140, f (0) = 51.02;
(c) f (5) − f (4.5) = 8.33;
(d) (1.4)3 ;
(e) 174.4%;
(f) 0.287.
WORKSHEET #18
1 A(3) = 40, 000, A(4) = 50, 000, B(3) = 15, 208.75, B(4) = 17, 490.06;
3 Sequence A’s increment is 10,000.
Sequence B’s multiplier is 1.15. Its proportionate change is 0.15.
4 A(k) = 10, 000 + 10, 000k; B(k) = 10, 000(1.15)k ;
5 B(25) is bigger;
7 B(30) = $391, 158.98;
10 C(k + 1) = 1.0752 C(k) = 1.155625C(k);
11 C(25) = $371, 897.46;
12 Condition C is best!
WORKSHEET #19
1 I. What is A(0) if A(18) = 40, 000?; II. What is B(3) if B(10) = 2048?
2 (a) divide by 2; (b) divide by 2; (c) divide by 22 ;
(d) divide by 24 ; (e) divide by 27 ;
3 16 million;
4 2 million;
5 multiply by (1.5)2 ; divide by (1.5)3 ;
6 Divide by (1.5)7 . Start with 119.865 million bacteria.
7 $34,293.55; $29,401.19;
178
A PPENDICES
8 A(0) = $10, 009.96;
9 $7432.18;
13 (a) $32.20; (b) 3.48 million; (c) $305.90; (d) $4523.49; (e) $134.87;
(f) 0.122 million = 122,000; (g) $350.09; (h) $256.
14 (a) 1.0379P dollars; (b) 0.9791P dollars;
(c) Each would have the same amount after 8 years. Mary would have more after 7 years.
WORKSHEET #20
1
(b) No, this doesn’t seem fair. Your friend invests more money than you do and should
receive more interest.
2
(b) multiplier=1.125, proportionate change = multiplier−1 = 0.125
3
(c) If we were earning 12.5% per year for 8 years, then we would get $20,526.28 at the
end of the 8 years. But we only get $16,000.
√
(a) p = 2 − 1 = 0.4142136, X = $11, 313.71
(b) You’ll earn $3,313.71 in interest; your friend will earn $4686.29 in interest.
(c) You’re each earning 41.42136% on your respective investments.
4
(a) A(1) = A(0) + A(0) · r OR A(1) = A(0) · (1 + r)
(b) A(2) = A(1) · (1 + r); A(2) = A(0) · (1 + r)2
(c) A(3) = A(2) · (1 + r); A(3) = A(0) · (1 + r)3
(d) multiplier=1 + r; proportionate change = multiplier −1 = r
(e) A(k) = A(0) · (1 + r)k
√
(f) r = 8 2 − 1 = 0.0905077. The effective annual interest rate is 9.05077%.
5
(a) A(0) = $1, 000; A(1) = $1, 060
(b) If r = 0.03, then A(1) = A(0) · (1 + 0.03)2 = 1, 000 · (1.03)2 = $1060.90.
√
(c) r = 1.06 − 1 = 0.029563
(d) X = 1, 000(1.06)1/2 = $1, 029.56
6 A(6.25) = 17, 000 · (1 + 0.055)6.25 = $23, 756.19
7
(a) 20 minutes =
(b)
1
3
of an hour; 40 minutes =
i. B(1) = 3 · B(0)
ii. B(2) = 32 · B(0)
iii. B(3) = 33 · B(0)
2
3
of an hour
A NSWERS TO S ELECTED E XERCISES
179
iv. B(k) = 3k · B(0)
(c) B
8
1
3
= 31/3 ·10 million = 14.422 million; B
2
3
= 32/3 ·10 million = 20.801 million
(a) $18,756.00;
(b) 20.668 million;
(c) $5259.95; $10,165.52;
(d) $3151.94;
(e) 1.768 million.
9 (a) 143.61 million; (b) 0.0247 million; (c) 58.65 million;
(d) 838%; (e) 4.36; (f) (1.75)5 .
10
(a) P (5) = k · m5 ; P (5.5) = k · m5.5 ;
(b) proportionate change = m0.5 − 1 or
√
(c) m = 3.61;
m − 1;
(d) 2376.099%;
(e) k =
7
3.613
(f) 48.013 million bacteria.
WORKSHEET #21
1 (a) additive; (b) A(t) = 4000 + 60t;
2 (a) I = rP ; (b) A(0) = P ; A(1) = P + rP ; A(2) = P + 2rP ; A(3) = P + 3rP ;
(c) A(t) = P + P rt = P (1 + rt);
3 (a) P = $2500, r = 0.01; (b) A(t) = 2500(1 + 0.01t); (c) (90 days = 0.25 years) A(0.25) =
2506.25;
4 (a) A(1) = $1551.75; (b) A(2) = $1605.29; (c) A(t) = 1500(1.0345)t ; (d) (54 months =
4.5 years); A(4.5) = $1747.34;
5 (a) A(2) = 500(1.00435)8 = $517.67; A(3) = 500(1.00435)12 = $526.73; A(t) =
500(1.00435)4t ; (b) (18 months = 1.5 years) A(1.5) = $513.19;
7 (a) P = 5000; r = 0.0247; n = 12; (b) A(t) = 5000 1 +
29
12
years) A
29
12
= $5307.22;
8 (a) P = 500; r = 0.0154; (b) A(t) = 500 1 +
9 (a) no! (b) no! (c) no!
0.0154
365
365t
0.0247
12
12t
; (c) (29 months =
; (c) A(1) = $507.76;
180
A PPENDICES
10 (b) less than 1 cent;
11 (a) P = 1000; r = 0.0311; (b) A(t) = 1000(e0.0311t ); (c) (30 months = 2.5 years) A(2.5) =
1080.85;
12 (a) $23,824.21;
(b) 1.197217;
(c) 1.51%;
(d) $0.85;
13 (a) $1239.00; (b) 6.89%; (c) $1.11; (d) $467.94;
(e) 0.0404; (f) $263.65; (g) $11,716.83;
.08 212
14 I is 2500(1 + 360
) ; II is 2500(1 +
IV is 2500(1 + 0.08)212/360 .
.08 424
) ;
720
III is 2500 · e(.08)(212/360) ;
WORKSHEET #22
1 Saucy’s is better, but just barely.
2
(a) i. I = A(1) − P = $30.68; ii. 6.136%;
(b) i. I = A(1) − P = $190.96; ii. 6.136%;
(c) i. I = A(1) − P = 0.06136P ; ii. 6.136%;
3 II. 6.144%; III. 6.051%
4 (a) I = P (er − 1); (b) (er − 1) × 100%;
5 5.985%;
WORKSHEET #23
1 (a) ln(20.08553692) = 3; (b) ln(22026.46576) = 10; (c) ln(2) = 0.6931471806;
(d) ln(10) = 2.302585093;
2 (a) 1.0986122789; (b) 5.991464547; (c) 6;
3 t = 17.06 years;
4 t = 5.31 years;
5 (a) A(t) = 3000(1.00435)4t ; (b) 6000 = 3000(1.00435)4t ; (c) t = 39.92 years;
6 (a) t = 44.52 years;
7 (a) (i) 6.39 years; (ii) 7.70 years; (iii) 11.55 years; (iv) 8.69 years;
(b) 2.79 years; (c) 15.024 years;
8
0.5r
(a) A: e
r
; B: 1 +
12
6
;
A NSWERS TO S ELECTED E XERCISES
(b) A: r = 0.09902; B: r = 0.09943;
√
(c) A: r = ln(s + 1); B: r = 12[ 12 s + 1 − 1];
(d) A: earns $1051.71 in interest in 2 years; takes 1.02 years to earn half that much;
B: earsn $1049.41 in interest in 2 years; takes 1.02 years to earn half that much;
9
(a) after 20 min: 2P million; after 40 min: 22 = 4P million;
after 60 min: 23 = 8P million;
(b) after 1 hr: 23 P million; after 2 hr: 26 P million; after 3 hr: 29 P million;
after t hr: 23t P million;
(c) t = 0.528 hours
181
182
A PPENDICES
G RAPHS AND TABLES
183
Graphs and Tables
This appendix contains copies of the graphs and tables in the book that you can tear out and include
in your homework.
184
A PPENDICES
G RAPHS AND TABLES
185
PROLOGUE
2
A
.z
...... ....
24 ... ....... .•..........
.. .. ..
.. .... ...
12 ... .... .....
....
.. ....
.. ....
.................................................................................................
... ..... a
..
.. 5 10 15 ...20
.... 25.......
−12 ...
... ....
..
......... ....... ....... ....... ....... ....... ....... ....... ..................
•...
−24 ..
.
..
B
y
.
............. ....... ....... ....... ....... ....... ....... ....... .....•........
2 ..
..... ........
.
..
.
.
.
.
..
..... ........
.
..
.
.
.
.
..
1 ......... ....... .................•.........
...
..
.. .. .....
....
.
...
..
..
...
..
..
...
..............................................................................................
..
1 2 3 4 5 x
C
.p
......
16 ..
(10, 12)......
.
12 ...
..........•
.. ........................
8 .. ....•
(2, 8)
.
4 ...
.
............................................................................................
q
2 4 6 8 10
186
A PPENDICES
G RAPHS AND TABLES
187
WORKSHEET #1
3
t
0
5
10
15
AT S
N/A
1.70
1.46
1.25
20
25
30
35
0.80
40
45
50
55
60
0.67
t = time in minutes, AT S = average trip speed in mpm
6
TI
0–5
5–10
10–15
∆D
8.5
6.1
4.1
AS
1.70
1.22
0.82
15–20
20–25
25–30
30–35
1.0
1.1
35–40
0.36
40–45
48
44
40
Distance (miles)
36
32
28
24
20
16
12
8
4
.....
..
..
.
..
.
..
..
..
.
.
...
..
..
.
.
..
..
..
.
.
..
..
..
.
.
..
..
..
.
.
..
...
.
..
.
.
.
...
......
.
.
.
..
.
.
.
.
.
.
..
...............
.
.
.
.
..
.
.
.
.
...
......
.
..
.
.
.
..
...
..
.
.
..
..
.
..
.
..
..
..
.
.
.. ..
.. ..
... ..
.. ..
.. ..
............................................................................................................................................................................. Time
10
20
30
40
50
60 (minutes)
50–55
4.0
0.32
T I=time interval, C=distance covered in interval, AS = average speed over interval
10
45–50
1.22
55–60
188
A PPENDICES
G RAPHS AND TABLES
13
d (in. miles)
....
24 ...
...
.
..
.
22 ...
.. ..
..
.
.
..
20 ...
.
.
..
..
.
.
.
.
18 .
..
..
.
.
..
16 ..
...
.
.
.
..
.
....
.
.
14 ...
.
.
.
.
.
...
..........
.
.
.
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.
.
.
.
.
.
12 .
..
...........
.
.
.
.
.
.
.
.
..
.
10 ..
.....
.
.
.
..
.
....
.
8 ...
.
..
..
..
.
.
6 ..
..
.
..
.
4 ... ...
.. ..
2 ... ...
..
.....................................................................................................................................................................................................
5 10 15 20 25 30 35 40 45 50 55 60
t (in minutes)
189
190
A PPENDICES
G RAPHS AND TABLES
191
WORKSHEET #2
3
t
0.0
0.5
1.0
1.5
O
0.0
3.6
6.1
7.7
2.0
2.5
3.0
3.5
4.0
10.46
11.46
4.5
5.0
15.56
I 0.0 2.0
t = hours since noon
O = total amount of water that flows out from noon to time t
I = total amount of water that flows in from noon to time t
4
24
Thousands of gallons
22
20
18
16
14
12
10
8
6
4
2
.
......
..
..
..
..
..
..
..
..
...
..
..
..
..
..
...
..
..
..
..
..
..
..
..
..
...
..
..
.............................................................................................................................................................................................................................................................................
1
2
3
4
5
6
Time (hours since noon)
5.5
6.0
192
A PPENDICES
G RAPHS AND TABLES
193
10
5000
4500
total water usage (gallons)
4000
3500
3000
2500
2000
1500
1000
500
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
hours after midnight
194
A PPENDICES
G RAPHS AND TABLES
195
WORKSHEET #3
2
q
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
p
N/A
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
TR
0
15
28
39
48
64
63
60
55
TC
0
4
8
28
32
36
40
44
P
0
35
32
27
20
24
48
0
3
.
......
..
..
..
..
..
..
..
..
..
..
..
..
..................................................................................................
.
8
.
.....
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..................................................................................................................................................
.....
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.................................................................................................................................................
60
196
A PPENDICES
G RAPHS AND TABLES
197
12
q
0
1
2
3
MR
15
13
11
9
MC
4
4
4
4
change in profit
11
5
4
4
5
4
6
4
7
4
8
4
9
4
10
−5
4
−9
11
12
13
14
15
4
4
4
4
4
198
A PPENDICES
G RAPHS AND TABLES
199
WORKSHEET #4
1
t
35
40
45
50
D
25.1
26.7
29.3
33.3
∆D
1.6
∆D
∆t
0.32
2
q
1
2
3
4
5
6
TR
15
28
39
48
55
60
∆T R
13
∆T R
∆q
5
q
1
2
3
4
5
6
7
8
9
10
11
12
TR
15
28
39
48
55
60
63
64
63
60
55
48
MR
13
11
9
7
5
3
1
AR
15
14
13
−1
−3
−5
−7
−9
7
60
50
40
30
20
10
.
...... T R
.............................................
..............
.........
.
..
.
.
.
.
.
.
.
........
....
..
.
.
.
.
.
.
........
...
..
.
.
.
.......
.
.
...
..
.
.
.
.
.
.
.
..
....
..
.....
.
..
.
..
..
....
.
..
.
..
..
....
.
.
..
....
..
.. ......
.. ....
.. ..
.....
..................................................................................................................................................................................................................................
q
2
4
6
8
10
12
200
A PPENDICES
G RAPHS AND TABLES
201
8
Course
Credits
Grade
Grade Points
Math 111
5
3.7
18.5
Psych 101
5
3.3
16.5
Engl 111
5
2.6
Music 116
2
4.0
10
Cr
15
30
45
60
QGP
60.0
52.5
39.0
QGPA
4.0
3.5
2.6
CGP
60.0
112.5
151.5
217.5
249.0
390.0
CGPA
4.0
3.75
3.37
2.42
2.075
2.17
1.9
75
1.4
90
1.1
105
120
1.0
1.1
11
.
......
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.............................................................................................................................................................................................................................................
135
150
165
21.0
28.5
39.0
1.4
1.9
2.6
180
3.5
202
A PPENDICES
G RAPHS AND TABLES
203
WORKSHEET #5
1
Time
5
10
15
20
25
30
35
40
45
50
55
60
Red
8.5
14.6
18.7
21.2
23.0
24.0
25.1
26.7
29.3
33.3
39.4
48.0
0
8.5
Purple
D(t)
3 .
.....
48 ...
...
.
..
.
..
R ...
.
.
40 .
...
.
..
.
.
.
..
....
.
.
.
.
.
.
.
.
32 .
..
.......
.
.
.
.
.
.
.
.
.
.
..
.
.
................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
24 ...
.
.
.
.
.
..
..............
.
.
.
.
.
.
.
..
.
.......
16 ...
.
.
.
.
..
....
.
.
..
.
.
8 ... ....
.. ..
....
.........................................................................................................................................................................................................................................................
5 10 15 20 25 30 35 40 45 50 55 60
204
A PPENDICES
G RAPHS AND TABLES
205
WORKSHEET #6
1
Amount (thousands of gallons)
30
27
24
21
18
15
12
9
6
3
.
......
..
..
..
....................
.
.
.
.
.
..
.
.
.
.
..
......
.
..
.
.
.
...
...
.
..
.
..
..
.
..
.
...
..
.
..
.
..
..
..
.
.
...
..
..
.
.
..
..
.
..
.
..
..
.
..
.
...
....
.
.
..
.
.
..
.....
.
.
.
..
.
.
.
.
.
.
..........................................................................................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
4
Reservoir
The (overall) average rate
of flow from midnight to
time t
The amount of water that
flows in from 2 hours to t
hours
The amount of water that
flows in from 5 hours to a
time h hours later
The (incremental) rate of
flow from a time t hours
to a time 21 -hour later
Functional Notation
Graph Language
206
A PPENDICES
G RAPHS AND TABLES
207
7
English
Functional Notation
Graph Language
A(4) − A(1)
the overall average rate
of flow after 5 hours
the slope of the secant
line from t = 1 to
t=1+h
the change in height
from t = 1 to t = T
A(4)
4
the incremental rate of
flow from t = 3 to a
time h hours later
8
y
16
14
y = f (x)
12
10
8
6
4
y = g(x)
2
x
1
2
3
4
5
6
7
8
9
10
208
A PPENDICES
G RAPHS AND TABLES
WORKSHEET #7
3
Amount (thousands of gallons)
30
27
24
21
18
15
12
9
6
3
8
3.2
2.8
Pounds
2.4
2.0
1.6
1.2
0.8
0.4
.
......
..
..
..
...................
.
.
.
.
..
.
.
.
.
.
..
.....
.
..
.
.
.
..
...
..
.
.
..
..
.
..
.
..
..
.
..
.
...
..
..
.
.
..
..
..
.
.
...
..
..
.
.
..
..
..
.
.
...
....
.
..
.
.
.
..
.....
.
.
.
.
..
.
.
.
.
.
.........................................................................................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
.
......
..
..................
.
.
.
.
..
.
B
.
.
.
..
.....
.
..
.
.
.
....
..............
.
.
.
.. ........
...
.
...........
.
...
.
.
............
.....
....
.
.. ....
.
.
.
.
.
.
.
.
............ ......
.. ....
....
.
.
....................
.......
.
..
.
........................................ ....................................................... A
..
..
..
..
..
..
..
..
.
............................................................................................................................................................................................................................
t
6 12 18 24 30 36 42 48 54 60 66 72
Hours
209
210
A PPENDICES
G RAPHS AND TABLES
211
9
70
60
50
TG
40
30
20
10
t
0
1
2
3
4
5
6
7
8
9 10 11 12
(Months)
10
Amount (thousands of gallons)
30
27
24
21
18
15
12
9
6
3
.
......
..
..
..
..............
.
.
.
.
.
.
.
.
.
..
.
..
........
.
.
.
.
.
..
.
..
....
.
..
.
..
.. ..
..
.
..
..
..
.
.
..
..
..
.
.
..
..
.
..
.
..
..
.
..
.
..
..
.
.
..
.
.
..
......
.
.
.
..
.
.
.
.
.
.
.....
........................................................................................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 11 12
Time (hours since midnight)
212
A PPENDICES
G RAPHS AND TABLES
213
11
y
16
14
y = f (x)
12
10
8
6
4
y = g(x)
2
x
1
2
3
4
5
6
7
8
9
10
214
A PPENDICES
G RAPHS AND TABLES
215
Cost (Dollars)
WORKSHEET
#8
......
480 ...
...
..
440 ...
.
.
..
..
...
.
.
400 ..
.. .
...
.
.
.. ...
360 ...
.
.
..
.. ..
..
.
.
.
.
.
320 .
... .. .
..
.
.
.
..
.
..
280 ..
....
.
.
.
.
.
..
.
TC.......
..
.
.
.
.
.
.
.
.
240 ....
.
.
..
..........
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
200 ...
...VC
.............................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...
.
.. ..
..........
.
.
.
.
.
.
160 ....
.
.
.
.
.
..
... .. .
.....
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
120 .. ....
... ... ... ... ... ...
.
.
.
.. ....
.
.
.
.
.
...
. ....
.
80 ....
.
.
.
..
. ...
.
.
.
.
.
40 . ...
.. .
.....................................................................................................................................................................................................................................................................................
..
50 100 150 200 250 300 350 400 450 500 550 600 650 700
Quantity
15
1200
1000
800
600
400
200
.....
..
..
..
..
..
..
.
..
.
TR
..
....................
.
.
.
.
.
.
.
.
.
.
..
.
..
........ ....
.
.
.
.
..
.
..
..VC
....
.
.
.
.
..
.
.
..
..
....
.
.
.
.
..
.
.
..
....
...
.
.
.
.
..
.
.
..
..
....
.
.
.
.
.. ...
.....
.. ..
.
.
.
.
.
.
.. ..
.
.... ...............
..................................................................................................................................................................
10 20 30 40 50 60 70 80 90 100
216
A PPENDICES
G RAPHS AND TABLES
217
Cost (Dollars)
WORKSHEET
#9
......
480 ...
...
..
440 ...
.
.
..
..
...
.
.
400 ..
.. .
...
.
.
.. ...
360 ...
.
.
..
.. ..
..
.
.
.
.
.
320 .
... .. .
..
.
.
.
..
.
..
280 ..
....
.
.
.
.
.
..
.
TC.......
..
.
.
.
.
.
.
.
.
240 ....
.
.
..
..........
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
200 ...
...VC
.............................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...
.
.. ..
..........
.
.
.
.
.
.
160 ....
.
.
.
.
.
..
... .. .
.....
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
120 .. ....
... ... ... ... ... ...
.
.
.
.. ....
.
.
.
.
.
...
. ....
.
80 ....
.
.
.
..
. ...
.
.
.
.
.
40 . ...
.. .
.....................................................................................................................................................................................................................................................................................
..
50 100 150 200 250 300 350 400 450 500 550 600 650 700
Quantity
218
A PPENDICES
13
2500
2250
2000
1750
TC
1500
dollars 1250
1000
750
500
250
0
0
200
400
600
800
1000
quantity (in items)
1200
1400
G RAPHS AND TABLES
219
14
9
8.5
8
7.5
7
6.5
6
5.5
dollars
5
per thumb4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
MC
AVC
0
0.5
1
1.5
2
2.5
3
3.5
hundreds of artificial thumbs
4
4.5
5
5.5
6
220
A PPENDICES
G RAPHS AND TABLES
WORKSHEET #10
240
2
220
200
thousands of dollars
180
160
140
120
100
80
60
40
20
0
221
Key Question I
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
.
.
........
.
.
.
.
.
.
.. .
.
........
.
.. . ..............
. ..
................
.
.......
.
.
.
.
.
.
.
.
........ ..
......... ...
..
.. .
..
.. .
.
0
1
2
3
4
5
6
Quantity (thousands of machines)
4
5
Key Question III
220
200
thousands of dollars
180
160
140
120
100
80
60
40
20
0
.
T R...
...
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
.
..
0
1
2
3
4
5
240
220
200
180
thousands of dollars
240
Key Question IV
160
140
120
100
80
60
40
20
0
6
Quantity (thousands of machines)
0
1
2
3
4
5
6
Quantity (thousands of machines)
222
A PPENDICES
G RAPHS AND TABLES
223
WORKSHEET #11
Distance (miles)
1
10
9
8
7
6
5
4
3
2
1
0
.
......
..
............................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
..
............
.
.
.
.
.
.
.
.
.
..
.
..
.........
.
.
.
.
.
.
.
.
..
.......
..
.
.
.
.
.
.
..
..
.......
.
.
.
.
.
..
....
..
.
.
.
..
.
..
....
.
.
.
.
..
..
....
.
.
.
.
..
.. ......
.. ....
.
............................................................................................................................................................................................................................................................................
0
2
4
6
8
10
12
14
16
18
20
Time (mins)
3
English
Graphs
Functional Notation
Formula or Numerical Value
the distance covered
by time t
the height of graph at
time t
D(t)
t − 0.025t2
the distance covered
by time 5
the height of the graph
at t = 5
D(5)
5 − 0.025 · 52 = 4.375 miles
the distance covered
by time 16
the height of the graph
at t = 9
224
A PPENDICES
G RAPHS AND TABLES
225
4
English
Graphs
Functional Notation
ave trip speed at time
t minutes
slope of diagonal line
to distance graph at t
D(t)
t
ave trip speed at time 6
minutes
slope of diagonal line
to distance graph at
t=6
D(6)
6
Formula or Numerical Value
t − 0.025t2
t
(6) − 0.025 · 62
= 0.85 mpm
6
ave trip speed at time
14 minutes
D(10)
10
8
English
Graphs
Functional Notation
average speed from
time t to time t + 5
minutes
slope of secant line
from t to t + 5
D(t + 5) − D(t)
5
average speed from
t = 4 to time t = 9
minutes
slope of secant line
from t = 4 to t = 9
D(9) − D(4)
5
slope of the secant
line from t = 6 to
t = 11
D(20) − D(15)
5
Formula or Numerical Value
226
A PPENDICES
G RAPHS AND TABLES
227
WORKSHEET #12
7
8
80
16
14
12
10
8
6
4
2
p
......
..
..
..
..
..
..
..
..
...
..
..
..
..
.................................................................................................
. q
2 4 6 8 10 12 14 16
70
60
50
40
30
20
10
$
.....
..
..
..
..
..
..
..
..
..
....
.
.
.
..
.
..
.... T C
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
..
.
..
....
.
.
.
.
..
....
..
.
.
.
.
..
..
....
.
.
.
..
.
..
....
.
.
.
..
.
.. ......
. ...
.............................................................................................................................................................................................................. q
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
228
A PPENDICES
G RAPHS AND TABLES
229
WORKSHEET #13
8
As Is
Answer
I.
Find the time at which the water in the red vat hits its lowest
level.
II.
Find the actual lowest value
the water level in the red vat
reaches.
III.
Find the time at which the
difference gauge reaches its
highest value.
IV.
Find the actual highest value
the difference gauge reaches.
Amount Shift
Time Shift
(R shifts up or down)
(R shifts left or right)
Change?
Change?
V. Find the first time at which the
water levels in the two vats are
the same.
Gallons
14
28
24
20
16
12
8
4
.
...... w
..
........ .........................
..
.. ... ...
.
.
.
.
.
.....
.. .......
R....
.
.. ... ....
.
..
.
.
....
.......
.
....
.
.
.
.... .........
..
.........
..
.
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.
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.
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.
.
.
.
........... . ....
..
...
..
..
...
..
..G
..
...
..
..............................................................................................................................................................
1 2 3 4 5 6 7 8 9 10 t
Hours
Answer
Answer
230
A PPENDICES
G RAPHS AND TABLES
231
WORKSHEET #14
1
3
8
6
4
2
5
y
......
.............
.. ..........
..........
..
..........
..
..
..........
..
..........
..
..........
..
..........
..
..........
..
.......
..
..
..
..
.............................................................................................................................................................
x
2
4
6
8
10 12
8
6
4
2
y ..
...... ....
.. ..
.. ..
.. ..
.. ..
...
..
..
....
..
......
..
............
..
................................
..
.....................
..
..
..
..
..............................................................................................................................................................
x
2
4
6
8
10 12
10
..... A
..
..
..
..
..
..
..
..
..
..
..
..
...
..
...
..
..
..
..
..
...
..
...
..
...
..
...
..
..
.............................................................................................................................................................
x
2
4
6
8
10 12
10
8
6
4
2
p
.......
..
.. ......
....... p = −0.6q + 10
..
.......
..
..
..
...........................................•....
........................................................................ ......
.......
........................................................................
.......
........................................................................
.......
........................................................................
........................................................................
.......
........................................................................
.......
........................................................................
.....
........................................................................
........................................................................
........................................................................
........................................................................
....................................
.................................................................................................................................................................................................
q
2
4
6
8
10 12
232
A PPENDICES
G RAPHS AND TABLES
12
233
15
10
8
6
4
2
p
.
...... ...
Price
.. ..
.. .......
.......... p = −0.6q + 10
..
..
... .........
..
.... ........
..
...... .......
..
........... ........
..
............................
..
...............................
..
8
..
.......
AC(q) = + 4
.. .......
q
..
....
.. AC ..
..
..
..
............................................................................................................................................................
q
2
4
6
8
10 12
10
8
6
4
2
p
.
...... ...
.. ..
Price
.. ....... ..........
..
..
... .........
..
.....................................................
............................................................................. ......
....
..
......
..............Profit
.....................
..
..
.
.. ..............................................................
.................................
..
.......
..
..
.......
..
.....
AC
..
..
..
..
..
............................................................................................................................................................
q
2
4
6
8
10 12
234
A PPENDICES
G RAPHS AND TABLES
235
WORKSHEET #15
Cost (Hundreds of Dollars)
4.80
4.00
3.20
2.40
1.60
0.80
.p
.......
...
...
...
...
...
...
..
...
...
..
...
TC
..
..
...
...
VC
..
...
..
...
...
..
...
.........................................................................................................................................
...
q
1
2
3
4
5
6
7
.
.....
.. ..
.
.
.. ..
.
.
... ...
.
.
.
.
.
.......... .....
.
.
.
.
.
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.
.
.
.
.
.
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.
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.
.
.
.
.
. ..
.........
.
.
.
.
.
.
.
.
..
... ... ... ...
.. .. .. .. ... ...
..
.
.
..
Quantity (Hundreds of Bags)
(Dollars per Bag)
2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
p
.......
...
MC
...
..
...
...
Breakeven
..
...
...
...
...
AC
...
...
...
...
...
..
........
...
AV C
•
...
...
..
..
•.............
..
.....
.....
..
..
...
Shutdown
...
..
..............................................................................................................................................
1
2
3
4
5
6
7 q
...
...
...
....
...
.
..
.
...
... .. ......
.....
..
... ....
.
... ..... ......... .... .......
..... . ..... .
... .....
... ........ .................. .........
........... ... ...........
...
...... ....
...
.
....
.
.
.
....... .....
.......
(Hundreds of Bags)
236
A PPENDICES
G RAPHS AND TABLES
WORKSHEET #16
50
10
40
30
20
10
.
...... d
..
..
...
..
..
..
..
..
..
..
..
..
...
..
..
..
..
.
.....................................................................................................................................
m
2
4
6
8
10
237
238
A PPENDICES
G RAPHS AND TABLES
239
WORKSHEET #20
1
Time
Supposed Value of Investment
4 years ago
$8,000
3 years ago
2 years ago
1 year ago
$11,000
now
$12,000
1 year from now
2 years from now
3 years from now
4 years from now
$16,000
Time
Supposed Value of Investment
4 years ago
$8,000
3 years ago
$8,000+8,000(0.125)=$9,000
2 years ago
$9,000+9,000(0.125)=$10,125
2
1 year ago
now
1 year from now
2 years from now
3 years from now
4 years from now
$12,814
240
A PPENDICES
G RAPHS AND TABLES
241
WORKSHEET #21
10
change in n
change in balance
from n = 1 to n = 2
$25.32
from n = 2 to n = 4
from n = 4 to n = 12
from n = 12 to n = 365
from n = 365 to n = 8760
from n = 8760 to n = 525, 600