MATH148
SCCC
Project 3
Marketing
Page 1
ACKNOWLEDGEMENTS
THIS SET OF MATERIALS IS A PRELIMARY WORK THAT IS BASED HEAVILY ON
MATERIALS CURRENTLY PUBLISHED IN ELECTRONIC FORM BY THE MATHEMATICAL
ASSOCIATION OF AMERICA (MAA). THE ORIGINAL MATERIALS, CONCEPTS AND IDEAS
FOR THE PROJECTS WERE DEVELOPED AT THE UNIVERSITY OF ARIZONA BY DR.
RICHARD B. THOMPSON (MATHEMATICS) AND DR. CHRISTOPHER G. LAMOUREUX
(FINANCE). THIS VERSION OF THE MATERIAL IS BASED ON THAT WORK, WITH
REVISIONS TO CONTENT AND ORDER OF MATERIALS. MANY OF THE EXAMPLES AND
EXPLANATIONS ARE COPIED FROM THEIR WORK AND SOME OF THE FILES USED ARE
FROM THEIR ORIGINAL SET OF MATERIALS. HOWEVER, MANY OF THE EXAMPLES AND
FILES ARE NEW TO THIS VERSION OF THE MATERIAL. THE MAIN DIFFERENCE IS NOT
SO MUCH IN CONTENT BUT THE FORM IN WHICH THE IDEAS, MATERIALS, AND
PROJECT CONCEPTS ARE PRESENTED. THIS VERSION HAS A MORE TRADITIONAL
TEXTBOOK FEEL THAN THE ELECTRONIC VERSION CURRENTLY BEING PUBLISHED BY
THE MAA.
THE CURRENT SET OF MATERIALS IS AN EVOLVING OUTGROWTH OF MEETINGS AND
WORK BY MR. GERALD WRIGHT, MR. JOHN TOUTONGHI, DR. DOUG SOLOWAN, AND
DR. LAWRENCE MORALES, ALL OF WHOM TEACH FOR SEATTLE CENTRAL
COMMUNITY COLLEGE.
Page 2
Table of Contents TABLE OF CONTENTS .............................................................................................................. 3
CHAPTER 1: BUSINESS BACKGROUND............................................................................... 5
SECTION 1.1: INTRODUCTION ...................................................................................................... 5
SECTION 1.2: THE CLASS PROJECT DATA ................................................................................... 6
SECTION 1.3: FOCUS ON THE PROJECT ........................................................................................ 8
PRACTICE PROBLEMS: A REVIEW OF ALGEBRA .......................................................................... 9
CHAPTER 2: GRAPHING & COST FUNCTIONS................................................................ 15
SECTION 2.1: INTRODUCTION .................................................................................................... 15
SECTION 2.2: USING GRAPHMATICA ......................................................................................... 19
SECTION 2.3 GRAPHING PIECEWISE FUNCTIONS ....................................................................... 23
SECTION 2.4: COST FUNCTIONS ................................................................................................. 26
SECTION 2.5: PIECEWISE COST FUNCTIONS .............................................................................. 29
SECTION 2.6: FOCUS ON THE PROJECT ...................................................................................... 32
CHAPTER 2 PRACTICE PROBLEMS ............................................................................................. 41
CHAPTER 3: TRENDLINES & DEMAND FUNCTIONS .................................................... 49
SECTION 3.1: INTRODUCTION .................................................................................................... 49
SECTION 3.2: CONSTRUCTING A LINEAR TRENDLINE ............................................................... 50
SECTION 3.3: THE LINEST COMMAND ..................................................................................... 54
SECTION 3.4: A REVIEW OF FUNCTION SHAPES ........................................................................ 59
SECTION 3.5: DEALING WITH UNITS .......................................................................................... 63
SECTION 3.6: DEMAND FUNCTIONS ........................................................................................... 64
SECTION 3.7: FOCUS ON THE PROJECT ...................................................................................... 68
CHAPTER 3 PRACTICE PROBLEMS ............................................................................................. 73
CHAPTER 4: REVENUE AND PROFIT FUNCTIONS ........................................................ 78
SECTION 4.1: THE REVENUE FUNCTION .................................................................................... 78
SECTION 4.2: PROFIT FUNCTIONS .............................................................................................. 84
SECTION 4.3: FOCUS ON THE PROJECT ...................................................................................... 89
CHAPTER 4 PRACTICE PROBLEMS ............................................................................................. 96
CHAPTER 5: LIMITS................................................................................................................ 99
SECTION 5.1: LIMITS .................................................................................................................. 99
SECTION 5.2: FINDING LIMITS WITH EXCEL ............................................................................ 102
CHAPTER 5 PRACTICE PROBLEMS ........................................................................................... 104
CHAPTER 6: RATES OF CHANGE AND MARGINAL ANALYSIS ............................... 106
SECTION 6.1: AVERAGE RATE OF CHANGE AND THE SECANT LINE........................................ 106
SECTION 6.2: INSTANTANEOUS RATE OF CHANGE AND THE TANGENT LINE.......................... 111
SECTION 6.3: MARGINAL ANALYSIS AND INSTANTANEOUS RATE OF CHANGE...................... 114
SECTION 6.4: FOCUS ON THE PROJECT .................................................................................... 123
CHAPTER 6 PRACTICE PROBLEMS ........................................................................................... 126
Page 3
CHAPTER 7: NUMERICAL DERIVATIVES ...................................................................... 133
SECTION 7.1: DERIVATIVES ..................................................................................................... 133
SECTION 7.2: COMPUTING NUMERICAL DERIVATIVES............................................................ 135
SECTION 7.3: FOCUS ON THE PROJECT .................................................................................... 137
CHAPTER 7 PRACTICE PROBLEMS ........................................................................................... 139
CHAPTER 8: SYMBOLIC DERIVATIVES AND GRAPHS OF DERIVATIVES ........... 141
SECTION 8.1: NOTATION AND BASIC RULES OF DERIVATIVES ............................................... 141
SECTION 8.2: POWER RULE ..................................................................................................... 144
SECTION 8.3: DERIVATIVES TERM-BY-TERM .......................................................................... 147
SECTION 8.4: GRAPHS OF DERIVATIVES .................................................................................. 151
SECTION 8.5: OPTIMIZATION AND APPLICATIONS ................................................................... 158
SECTION 8.6: FOCUS ON THE PROJECT .................................................................................... 167
CHAPTER 8 PRACTICE PROBLEMS ........................................................................................... 172
CHAPTER 9: EXCEL'S SOLVER (OPTIONAL)................................................................. 177
SECTION 9.1: INTRODUCTION .................................................................................................. 177
SECTION 9.2: USING THE SOLVER FEATURE............................................................................ 177
SECTION 9.3: FOCUS ON THE PROJECT .................................................................................... 184
CHAPTER 9 PRACTICE PROBLEMS ........................................................................................... 186
CHAPTER 10: INTEGRATION ............................................................................................. 189
SECTION 10.1: AREA UNDER A CURVE.................................................................................... 189
SECTION 10.2: CONSUMER SURPLUS ....................................................................................... 191
SECTION 10.3: ESTIMATING AREA UNDER A CURVE .............................................................. 196
SECTION 10.4: INTEGRALS ....................................................................................................... 203
SECTION 10.5: BASIC INTEGRAL RULES .................................................................................. 204
SECTION 10.6: THE FUNDAMENTAL THEOREM OF CALCULUS ................................................ 207
SECTION 10.7: INTEGRATION WITH GRAPHMATICA ................................................................ 211
SECTION 10.8: FOCUS ON THE PROJECT .................................................................................. 214
CHAPTER 10 PRACTICE PROBLEMS ......................................................................................... 217
INDEX ........................................................................................................................................ 226
Page 4
Chapter 1: Business Background Reading Questions 1.) Describe the Team Project in your own words. What is it that you'll be doing?
2.) In the project, state TWO key questions we want to answer about profit?
3.) Describe what Table 1, titled “Test Markets” tells you. Be specific. For example, what key
relation ship does it describe?
Section 1.1: Introduction A company makes a positive profit when revenues from selling a product exceed the costs of its
production. If a company is a monopoly, then the amount of its product that the company
attempts to sell will affect the price that it can charge for the product.
This contrasts with a perfect competitor, who cannot affect a good’s price by changing the
amount that it attempts to sell. As an example of a perfect competitor, imagine the operator of a
small farm. If she were to plant wheat instead of soybeans, there would be no effect on the price
of either commodity.
Pure monopolies are rare. The simplest examples are natural monopolies, such as public
utilities, which are usually regulated by governmental agencies. Sometimes a new technology
can give partial monopoly power to a company. Advertising also plays a role in enhancing a
company’s monopolistic power by differentiating the company’s product from similar offerings
of its competitors.
A monopolist generally has a down-ward sloping demand curve. This is the graph of the
demand function, that relates the price which a company can charge for each unit of its product
to the quantity of the product that it sells. No company can know its demand function exactly,
but it can estimate the general properties of the function. The overall slope of a demand curve
has business implications. For example, a medicinal drug might have a steeply dropping demand
curve, whereas a new computer accessory seller might have a slowly dropping demand curve.
In this project, we let q be the number of units of a good that are sold, and we let p be the price
per unit at which that good is sold. The demand function relates these variables. A company’s
revenue function is given by the quantity sold multiplied by the price we charge per unit. In
symbols, this is R = q ⋅ p . If a company sells a small amount of its good (small q), it can sell each
unit for a high price (large p), but the total revenue may be small. If the company sells a very
large quantity of its good (large q), then the unit price will be very low (small p). This will also
result in low total revenue. In a monopolistic situation, the company wants to use its demand
function to determine values for q and p that will produce the largest possible total revenue.
When making goods, companies incur costs. The total cost function of production arises from
the production process, and can be estimated by engineers and managers. There are fixed costs of
production, which are incurred even if the company produces no units of its good. Variable costs
are incurred based on how many items are produced.
Chapter 1 – Page 5
The profit that a company makes is given by its revenue minus its costs. All companies confront
the basic problem of determining the quantity of goods that should be produced and sold in
order to maximize profit. In this project, we use mathematics to address such a problem.
Section 1.2: The Class Project Data For this project, we will be using the following scenario and data to work through the
mathematics necessary to complete the team project.
Save-it-All!, Inc. has just developed and patented a new type of computer drive, the SXL. The
new drive features reliability, compact size, and the ability to store a large amount of information.
Under the conditions of their patent, Save-it-All! has the exclusive right to produce and market
the new technology during the coming three years. This gives the company temporary
monopolistic power.
The management team at Save-it-All! wants to know how to price the SXL in such a way that it
will produce the maximum profit during the coming year. The management team also wants to
know how many of the drives that they can expect to sell, and how much profit they might hope
to realize from those sales.
The Marketing Department at Save-it-All! has estimated that there are 120 million potential
customers in the overall market for the coming year. They studied six test markets to determine
the fraction of the potential buyers who would actually purchase the SXL, at various price levels.
The results of these test markets are shown in the following table and in the Excel file Marketing
Data.xls.
Table 1
Test Markets
Market
Number
Market
Size
Price
1
2
3
4
5
6
1,956,000
1,044,000
492,000
1,512,000
1,104,000
1,224,000
$119.95
$129.95
$139.95
$154.95
$169.95
$179.95
Projected Yearly
Sales
(number of drives)
14,563
7,143
3,179
8,404
5,297
4,573
Units:
Here are the units we'll use in the Project. We will need to pay close attention to them throughout
this project:
Prices for individual drives are given in dollars.
Projected quantities of drives in the test markets are actual numbers of drives.
Projected quantities of drives in the national market are given in thousands of drives.
Revenue, Costs, and Profits are in millions dollars.
Chapter 1 – Page 6
Past experience leads the Marketing Department to assume that the SXL drives will have a
quadratic demand function.
Obviously, the costs involved in manufacturing the SXL drives will play a major role in
determining the appropriate price for the new product. The Production Department at Save-itAll! estimates that there will be fixed overhead costs of $21,600,000 during the coming year.
The engineers at Save-it-All! made the following estimates for the production costs of several
different quantities of drives.
•
•
•
The first 500,000 drives will have production costs of $115 per drive. We will call
500,000 the first “economy of scales” point.
The next 600,000 drives will cost $100 per drive to produce. This means all units
between 600,000 and 1,100,000 cast $100 per item. We will call 1,100,000 the second
“economy of scales” point.
All drives after the first 1,100,000 can be produced for $90 per drive.
Notice that as more units are produced, the cost per unit decreases. This is a simple example of
what is called “economies of scale.”1
We will use all of this available information to analyze the pricing and production of SXL drives.
Our ultimate goal is to answer the following questions:
Questions to Answer
1.
2.
3.
4.
What price should Save-it-All! put on the drives, in order to achieve the maximum profit?
How many drives might they expect to sell at the optimal price?
What maximum profit can be expected from sales of the SXL?
How sensitive is profit to changes from the optimal quantity of drives, as found in Question 2?
(That is, as you increase or decrease quantity sold from the optimal quantity, how is profit
affected?)
5. What is the consumer surplus if profit is maximized?
Save-it-All! is considering its options in terms of advertising and of putting more capital into
its business. We are to analyze the following questions.
6. What profit could Save-it-All! expect, if they price the drives at $154.49?
7. How much should Save-it-All! pay for an advertising campaign that would increase demand
for the SXL drives by 10% at all price levels?
8. How would the 10% increase in demand effect the optimal price of the drives?
9. Would it be wise for Save-it-All! to put $4,000,000 into training and streamlining which
would reduce the variable production costs by 7% for the coming year?
1
There are various ways to utilize economies of scale, including bulk purchasing/production, increasing
specialization of managers, lower interest rates when borrowing, and effective marketing.
Chapter 1 – Page 7
Section 1.3: Focus on the Project Each team will receive an Excel file containing test market data and production cost estimates for
a new product. This data applies to the coming sales year for the product. Assuming that you
have a temporary monopoly for the sale of all similar items, your team is to answer (at least) the
following questions:
1.
2.
3.
4.
5.
What price should be put on the new good, in order to achieve the maximum profit?
How many items might you expect to sell during the coming year at the optimal price?
What yearly maximum profit can be expected from sales of the new product?
How sensitive is profit to changes away from the optimal quantity of items, as found in
Question 2?
What is the consumer surplus if profit is maximized?
Your Excel file will also contain additional questions (like those numbered 6 to 9 in the previous
box) on changes in demand and capital investments. Consult with your instructor about which
questions you must answer for your oral and written reports.
Chapter 1 – Page 8
Practice Problems: A Review of Algebra In this project it will be necessary to use a variety of ideas from algebra that you have seen
before. The following problems are intended as review and should be exercises you can complete
with relative ease. If there are questions that you do not remember how to answer, we suggest you
take a little time to crack open your old algebra text (or grab one from the library) and review as
necessary.
1.
Solve for x:
3( x + 2) = 37
2.
Solve for w:
0.8w − 2.6 = 1.4w + .3
3.
Solve for t:
0.25t + 0.1(t − 4) = 11.6
4.
Solve for z:
3
(z + 1) = 1
2
4
5.
A boutique carries a line of jewelry made by a group of local artists. If the owner of the
boutique charges p dollars for a pair of earrings, he finds he can sell 200 − 5 p pairs per
month. On the other hand, the artists will provide the boutique with 56 + 3 p pairs of
earrings when they charge p dollars per pair. What price should the boutique owner charge
so that the demand for earrings will equal the supply?
6.
Last year, Pinwheel Industries introduced a new calculator. It cost $2000 to develop the
calculator and $20 to manufacture each one.
Write an equation that expresses the total cost, C, in terms of the number, n, of calculators
produced.
Graph the equation, using proper scales and axes labels.
How many calculators can be produced for $10,000?
a.
b.
c.
7.
a.
b.
c.
The world's oil reserves were 1660 billion barrels in 1976: total annual consumption is 20
billion barrels.
Write an equation that expresses the remaining oil reserves, R, in terms of time, t (years
since 1976).
Find the intercepts and graph the equation.
What is the significance of the intercepts in this context? In other words, what do they
mean?
8.
a.
b.
Find the slope of the line that goes through each pair of given points:
(−1,4) and (3,−2)
(5,0) and (2,−6)
9.
a.
b.
Find the equation of the line that goes through each pair of given points:
(−3,−1) and (2,4)
(3,5,) and (8,15)
Chapter 1 – Page 9
10.
The table shows the amount of oil, B (in thousands of barrels), left in a tanker t minutes
after it hits an iceberg and springs a leak.
0
10 20 30
t
B 800 750 700 650
a.
b.
c.
Write a linear equation that expresses B in terms of t.
Graph the equation.
Give the slope of the graph, including units, and explain the meaning of the slope in terms
of the oil leak.
11.
A business finds there is a linear relationship between the number of shirts it can sell and
the price per shirt. In particular, 20,000 shirts can be sold at $19 each, and 2,000 shirts can
be sold at $55 each. Write a linear equation that gives the price, p, that they can charge as a
function of the number of shirts, n.
12.
Find the equation of the line shown.
6
y
3
x
-6
-3
0
3
Chapter 1 – Page 10
6
9
13.
Find the equation of the line shown:
y
8
x
-24
-16
-8
0
For the following questions use the quadratic formula to find the solutions of the equation, if
there are any. Recall that the quadratic formula gives you the solutions to any equation of the
2
form ax + bx + c = 0 , provided solutions exist. The quadratic formula says:
x=
− b ± b 2 − 4ac
2a
14.
2 x 2 + 5x + 1 = 0
15.
5y 2 + 6y = 2
16.
2r 2 − 7 r + 5 = 0
17.
2 x 2 + 30 = 7 x
18.
4.42 x 2 − 10.14 x + 3.79 = 0
19.
If F (t ) = 1 + 4t 2 , find F (0) and F (−3) .
20.
If f ( x) = 2 − 3 x , find f (2) + f (3) and f ( 2 + 3) .
21.
The total number of barrels of oil pumped by the AQ oil company is given by the function:
N (t ) = 2000 + 500t
where N is the number of barrels of oil t days after a new well is opened. Evaluate N (10)
and explain what it means.
Chapter 1 – Page 11
For the next two problems,, use the graph of g (x) , shown below:
12
y
8
4
x
-2
-1
0
22.
Find the values of g (1) and g (−1) .
23.
For what value(s) of x is g ( x) = 4 .
1
Chapter 1 – Page 12
2
Answers to Selected Problems 1.
x=31/3
2.
w =−4.833
3.
t =34.286
4.
z = −1/3
5.
18
6.
a.
b.
c.
7.
a.
b.
c.
C = 2000 + 20n
n = 400
R = 1660 − 20t
R intercept is 1660, t intercept is 83.
8.
a.
b.
−3/2
2
9.
a.
b.
y = x+2
y = 2x−1
10.
a.
b.
c.
B = −5t + 800
The slope is −5 thousand barrels per minute and it means that the amount of oil left
is decreasing by 5,000 barrels per minute.
11.
p = −.002n+59
12.
y = −.5x + 3
13.
y =x/3 + 8
14.
x = −0.219, −2.281
15.
y =0.272, −1.472
16.
r = 2.5, 1
Chapter 1 – Page 13
17.
Solutions do not exist.
18.
x = 0.470, 1.824
19.
F(0) = 1; F(−3) = 37
20.
f(2)+f(3)= −11; f(2+3)=−13
21.
10 days after a well opens, the well has produced a total of 7000 barrels of oil.
22.
g(1) = 12; g(−1) = 8
23.
At about x = 1.5 and x = −1.25
Chapter 1 – Page 14
Chapter 2: Graphing & Cost Functions Reading Questions 1.) What is a Cost function? What does it tell you? Be specific. You should be able to do this
in one or two sentences, at the most.
2.) Suppose C (q) = 400 + 3q , where q is items produced and C (q ) is in dollars.
a.
What are the fixed costs?
b.
What is the value of C (75) ? Show your work and in your own words, explain what the
result means.
3.) In general, what is meant by a "piecewise function” in mathematics? Use ONE sentence.
4.) Explain why need to use a piecewise function in this project.
⎧5 + 1.5x if x < 0
⎪
5.) Find T (3.5) and T (−7) if T ( x) = ⎨5-x if 0 ≤ x < 5 . Show all work.
⎪2 x-10 if x ≥ 5
⎩
6.) Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 2.1: Introduction We assume you have a solid working knowledge of how Excel works. Having said that, we'll
start with a simple example of how to graph functions using Excel.
Example 1 Let’s graph f ( x) = x 2 from -10 to +10 on the x-axis and include 25 subintervals total in our
graph.
Part (1)
We’ll say that the interval we are graphing is [-10,10], which has the general form [a,b]. In
these examples and the spreadsheets we will use, a is the leftmost endpoint to be graphed
and b is the rightmost.
If we take the value b-a, we get the total length of the domain. In this case
a − b = 10 − (−10) = 20 . So our interval is 20 units long (-10 to 0 and then 0 to 10).
Our next step is to take that entire interval and compute the length of each subinterval. The
length of this space will be called the subinterval length and we will use the symbol Δx to
represent it. We simply need to take the length of the interval and divide by how many
subintervals we have to get Δx . In this case, we have:
Δx =
10 − (−10) 20
=
= 0.8
25
25
(Technically, if we have 25 subintervals we need 26 points. You always need one more
point than you have subintervals {why?}. When graphing functions, we will divide by the
number of subintervals, which is always one less than the number of points needed.)
Chapter 2 – Page 15
The top of the file Parabola.xls shows how these have been entered and computed in Excel.
You should open this file and examine the formulas and entries in cells B1 through B5
before continuing. A snapshot is show below:
The following picture shows which formulas were used in each cell. (You can use CTRL-`
to toggle between Data view and Formula view...it’s a handy feature to know about.)
Part (2)
In this next part we need to generate a list of ( x, y ) points to graph. We start with the left
endpoint (a) which is -10 in this example. That’s our first x value. The next x value is the
previous one PLUS the length of each subinterval. In this example, it would be −10 + 0.8 =
−9.2. The third x value will be −9.2 + 0.8 = −8.4 and so on. We certainly don’t want to
compute and type these in by hand so we will have Excel do it.
In Cell B8, the value of a is entered. In Cell B9, we need to take Cell B8 and add Δx = 0.8
to it. We could enter into Cell B9 the formula =B8+B5. However, when we copy this
down, Excel will assume we want relative references and it will change B5 to B6, which is
blank. This is an example of when we need to utilize absolute references to get the results
we desire. So what we put into Cell B9 is =B8+$B$5. Recall that the dollar signs lock us
into using Column B and Row 5. We can then copy this down to get to +10 for the last x
value. Open the file Parabola.xls and see how this was done. Make sure you understand
how the x values have been generated and note that there are exactly 26 points (25+1), as
discussed before.
We can now easily generate the y values. In Cell C8, the simply formula =B8^2 is entered
to take the x value in Cell B8 and square it. Don’t forget, the function is f ( x) = x 2 , so this
makes sense. We now use Ctrl−D to Copy Down our formula and then insert a Scatter
graph with no connection lines…just the points. This gives the following graph, making no
other special changes:
Chapter 2 – Page 16
Of course, we could pretty this up, but the main thing we note here is that we usually don’t
graph functions with dots. We usually prefer nice smooth curves or lines without the dots.
Let’s change the chart type to a smooth line graph with no data points. The choice is
highlighted below:
This gives us the following graph, which we can pretty up with colors, labels, etc. as we see
fit. See the sheet “Chart2” of the Parabola.xls file for a nice picture of the parabola.
Chapter 2 – Page 17
That can be considered a lot of work to draw a simple function. Excel is good at number
crunching, but other tools do a better job of graphing functions and will be useful for us here.
Chapter 2 – Page 18
Section 2.2: Using Graphmatica Graphmatica is an example of shareware software that does a great job of graphing functions. It
has the added benefit of being easy to use as well.2 Here’s a screen shot of the basic software.
Equation bar
You’ll notice a white box near the top of the screen, which is the Equation bar. This is where all
of your equations are typed. Graphmatica can display several equations at a time, which will
come in handy for us. Off to the right you see the “Point Tables” region, which automatically
substitutes values of x into your function. The easiest way to see how it works is to input an
equation.
2
At SCCC, Graphmatica is on the student network. Ask your instructor for where to find it. We’ll just
handle basic functionality in this text. Your instructor may show you more advanced features in class.
Chapter 2 – Page 19
Let’s input f ( x) = x 2 as in the previous example. One key rule in Graphmatica is that it only
recognizes x and y as the main variables. Therefore, we have to re-write our function as y = x 2
when we put it into the Equation bar:
Notice that Graphmatica uses the same notation as Excel does for exponents. When you hit Enter
on your keyboard, the graph will immediately be drawn for you:
Chapter 2 – Page 20
Notice that the Point Table is automatically filled in for you with several values. We can
eliminate the white space below the graph by zooming into the area we want. With the mouse, we
can click and hold to draw a box around the region where we want to zoom. It will look like this:
Then, we click on the Zoom In icon on the tool bar:
Chapter 2 – Page 21
When we zoom in, we see the following:
The Point Table automatically adjusts its values when you zoom in or zoom out, as you can see
above. Obviously, this is much easier than using Excel. Not only that, but you can take any
picture in Graphmatica, copy it, and paste it into any Word or PowerPoint document. You can
find the Copy commands in the Edit menu:
Chapter 2 – Page 22
This gives you a quick intro to how the software works. It is worth the time to play with it and get
comfortable with its features. We’ll see more of them in a moment.
Section 2.3 Graphing Piecewise Functions In many situations, the type of function you have changes depending on the value of x. For
example, if you order 100 or fewer items, the price might be $4 each. However, if you order 101
or more items, you get the “bulk rate,” perhaps $3.50 per item thereafter. This is an example of a
piecewise function. It gets its name from the fact that the function does not have just one
equation, but instead it must be specified in pieces. In this example, we’ll let x be the number of
items, and C(x) is the cost. If x is less than or equal to 100, then C(x) = 4x. But, if x is greater than
100, then C(x) = 400 + 3.5(x – 100). The 400 comes from the fact that the first 100 items cost $4
each, for a total of $400. The (x – 100) comes from the fact that only the number of items after
the first 100 cost $3.50. In order to simplify this written expression of C(x), we would write:
⎧4 x if x ≤ 100
C ( x) = ⎨
⎩400 + 3.5(x − 100 ) if x > 100
Let's try a few examples to make sure we have the idea.
Example 2 Let f (x) be the following piecewise function:
⎧ x + 1 if x < 0
⎪
f ( x) = ⎨− 2 x + 7 if 0 ≤ x ≤ 2
⎪ x 2 if x > 2
⎩
What is the value of f (−1) ?
Solution:
Since −1 < 0, then f ( x) = x + 1 . So, f ( −1) = −1 + 1 = 0 .
Answer: f ( −1) = 0 . ■
Example 3 Let f (x) be given as in the previous example. What is f (3) ?
Solution:
Since 3 > 2 , f ( x) = x 2 . f (3) = 32 = 9.
Answer: f (3) = 9 . ■
Chapter 2 – Page 23
Example 4 Let f (x) be given as in the previous example. What is f (1) ?
Solution:
Since 0 ≤ 1 ≤ 2 , f ( x) = 2 x + 7 , so f (1) = −2(1) + 7 = 5 .
Answer: f (1) = 5 . ■
Now, we know a function usually has a graph associated with it. Let's consider once again the
following function:
⎧ x + 1 if x < 0
⎪
f ( x) = ⎨− 2 x + 7 if 0 ≤ x ≤ 2
⎪ x 2 if x > 2
⎩
Because this function has three pieces or parts, it will have three different parts on its graph. The
first two will be lines and the third will be a parabola. Let's first draw all three functions on the
same set of axes, and ignore the restrictions on the values of x that can be used. This is what we
get.
y
10
5
y = x +1
y = x2
x
-4
-2
0
2
4
y = −2 x + 7
-5
Note: To generate these graphs, I entered each one individually into Graphmatica. The software
graphs the entire functions because it does not know that each is restricted on a particular domain.
The Equation box looks something like this when you click on the drop down arrow on the right
side of the Equation box:
Chapter 2 – Page 24
We can force Graphmatica to graph each function only on a specified domain. To do so, we use
the { } symbols to modify the domain. For example, to graph y = −2 x + 7 on the interval from
x = 0 to x = 2 , we would type:
y = -2x+7 {0,2}
If you don’t have an endpoint for your domain, you leave the corresponding space blank. So, for
example, to graph y = x 2 from x = 2 forward, we type:
y = x^2 {2,}
And, to graph y = x + 1 for x < 0 , we would enter:
y = x+1 {,0}
When we enter only these equations, with the domain modifiers, the graph looks like this:
y
10
x^2
-2x+7
5
x
-4
-2
0
2
4
x+1
-5
As you can see, this graph is literally in pieces. Hence the name, “piecewise function!”
Excel can do this, of course, but it’s a lot more work and it’s a lot more tedious, as you can
probably imagine. (Challenge: Try to use Excel to graph this piecewise function.)
Chapter 2 – Page 25
Example 5 ⎧ x 2 if x < −1
⎪
Graph the function f ( x) = ⎨ x + 2 if − 1 ≤ x ≤ 1
⎪4-x if x > 1
⎩
over the interval [−5,5] using 100 subintervals.
Solution:
The graph can be obtained by entering the correct equations with domain modifiers, one
at a time. Here are the three equations we need:
y = x^2 {,-1}
y = x+2 {-1,1}
y = 4-x {1,}
This will produce the following graph. Notice how the endpoints “link up” with each
other, unlike the previous example.
y
20
10
x
-6
-4
-2
0
2
4
6
You should try to reproduce this graph yourself using the software. ■
Section 2.4: Cost Functions A Cost function (sometimes called the Total Cost function) is an equation or formula that gives
information about the total costs required to produce a certain number of units of some item. We
will usually let the number of units be specified by the variable q. We will also let the Cost
function be designated by C (q ). A Cost function is often made up of two different pieces. The
first one is called the fixed costs, or the overhead costs. These are costs that do not change if the
number of units produced changes. They could be expenses such as buying production
Chapter 2 – Page 26
equipment, initial setup fees, or other costs that are associated with starting a business or a new
product line. There is also what we call variable costs. These are costs that change as we
produce more units. These costs may be associated with the process of actually producing an
item, such as the cost of raw materials needed to produce them, the labor needed to support their
production, or the utility costs required in the production process. A typical Cost function will
have the following form:
The Cost Function
C (q) = Variable Costs + Fixed Costs
Example 6
A company is studying a new product line. They have fixed costs of $2 million and each item
produced will cost $12.50. Find the total Cost function for the new product.
Solution:
The fixed costs in this situation are given as $2 million. The variable costs are $12.50.
Answer: The total cost function is C (q) = 12.5q + 2,000,000 . ■
Example 7 The Cost function for an item being produced by a company is given by the equation:
C (q) = 34.5q + 100,000
q is number of units produced and C (q ) is the total cost produce q units in dollars.
a.
What are the fixed costs and what are the variable costs?
b.
How much will cost to produce 3000 items?
c.
How many items can be produced with half a million dollars?
Solution:
Part a.
Answer a: The fixed costs are $100,000. The variable costs are $34.50 per item.
Part b.
We are given that q = 3000 . We only need to plug that into the Cost function:
C (3000) = 34.50(3000) + 100,000
= 103,500 + 100,000
= 203,500
Answer b: It will cost $203,500 to produce 3000 units.
Chapter 2 – Page 27
Part c.
In this question we are given the money available to us to spend on costs. We know that
C (q) = 500,000 and we want to find the value of q.
500,000 = 34.5q + 100,000
500,000 − 100,000 = 34.5q + 100,000 − 100,000
400,000 = 34.5q
400,000 34.5q
=
34.5
34.5
11594.20 = q
Since it’s not usually possible to produce 0.20 of an item, we’ll simply round down to
get our final result.
Answer c: We can produce about 11,594 units with $500,000. ■
Example 8 A company estimates it can produce 10,000 items for $620,000 and 20,000 units for $890,000.
Assuming a linear rate of growth for total costs, find the total Cost function for this product and
determine the fixed costs for this product.
Solution:
We are given two sets of data. When q = 10,000, C (q) = $620,000 . When q = 20,000,
C (q) = $890,000 . Because we are told we have linear growth, we need to find the
equation of the line that goes through these two points. We start by finding the slope of
the line, m:
m=
C ( q2 ) − C (q1 )
q2 − q1
890,000 − 620,000
20,000 − 10,000
270,000
=
10,000
= 27
=
This means that each unit costs $27 to produce. Another way to say this is that our costs
are increasing by $27 per unit. We can now find the rest of the equation which will be
of the form y = mx + b .
Chapter 2 – Page 28
The equation of any line, in terms of
x and y.
This is the same equation in terms of
our current variables.
Since we know a point on the line,
we can substitute it into our line's
equation and then…
y = mx + b
C (q) = mq + C0
620,000 = 27(10,000) + C0
620,000 = 270,000 + C0
620,000 − 270,000 = C0
…Solve for C0
350,000 = C0
C (q ) = 27 q + 350,000
Now write C (q) .
Answer: This is our total cost function: C (q ) = 27 q + 350,000 Notice that when we
substitute q = 0, we get total costs of $350,000. This is our value for fixed costs. ■
Section 2.5: Piecewise Cost Functions Since the cost to produce an item may depend on how many we produce, piecewise cost functions
are important tools for us in this project. Some simple examples will illustrate how.
The next few examples use the following scenario:
Suppose that a product has a fixed cost of $20,000,000. The variable costs are…
$45 per unit for the first 400,000 units
$40 per unit for the rest
Example 9 How much does it cost to make 100,000 units?
Solution: All 100,000 units will cost $45 each. So the total cost is:
Total cost = Fixed cost + cost of 100,000 units
= 20,000,000 + (45)(100,000)
= 20,000,000 + 4,500,000
= 24,500,000
Answer: It will cost $24,500,000 to produce 100,000 units. ■
Chapter 2 – Page 29
Example 10 How much does it cost to make 600,000 units?
Solution: The first 400,000 units will cost $45 each, but after that (the remaining
200,000 units) the cost decreases to $40 per unit.
Total cost = Fixed cost + cost of first 400,000 units + cost of remaining units
= 20,000,000 + (45)(400,000) + (40)(200,000)
= 20,000,000 + 18,000,000 + 8,000,000
= 46,000,000
Answer: It will cost $46,000,000 to make 600,000 units. ■
Example 11 Build a piecewise function for the cost of making q units.
Solution: This must be separated into two situations:
When less than 400,000 units are produced (in other words, q ≤ 400,000).
When more than 400,000 units are produced (in other words, q > 400,000).
Part 1: When less than 400,000 units are produced, then
Total cost = Fixed cost + cost of q units
= 20,000,000 + 45q
Part 2: When more than 400,000 units are produced, then
Total cost = Fixed cost + Cost of first 400,000 units + Cost of remaining units
= 20,000,000 + (45)(400,000) + (40)(q – 400,000)
Note that we multiply 40 by (q – 400,000), because we’re only interested in the number
of units remaining (after the first 400,000 were already made).
Next, we’ll use some algebra to simplify the expression:
20,000,000 + (45)(400,000) + (40)(q – 400,000)
= 20,000,000 + 18,000,000 + 40q – 16,000,000
= 22,000,000 + 40q
To summarize:
If q ≤ 400,000, then the function is 20,000,000 + 45q.
If q > 400,000, then the function is 22,000,000 + 40q.
Chapter 2 – Page 30
Answer: This is written in the following way:
⎧45q + 20,000,000 if q ≤ 400,000
C (q) = ⎨
⎩40q + 22,000,000 if q > 400,000
■
Chapter 2 – Page 31
Section 2.6: Focus on the Project How can graphing help us with our Project? Recall that we have the following information about
how much our computer drives will cost to produce:
The first 500,000 drives will have production costs of $115 per drive. The next 600,000
drives will cost $100 per drive to produce. All drives after the first 1,100,000 can be
produced for $90 per drive.
We also know that before producing any units, the company will have costs of $21.6
million dollars. These were called fixed overhead costs.
So, if q is the quantity of drives, and C (q ) is the total cost to produce q drives, we can use this
information to build a piecewise function that gives us total costs at any level or production, q.
Important note: Later in the project we will need to change our units to make them a little more
friendly for graphing, but for now we will simply use the "raw" units to practice building a
piecewise function from our data. There are three parts to the piecewise Cost function.
Example 12 Build the piecewise Cost function for the class project data:
Part 1: When less than 500,000 units are produced at $115 each:
Let's compute what we know, being careful to keep our units consistent.
If 500,000 drives or fewer are produced, then the cost is $115 each.
In terms of a function or expression we could say that C ( q) = 21,600,000 + 115q if
q ≤ 500,000 . This is the first part of our piece wise function:
⎧115q + 21,600,000 if q ≤ 500,000
⎪
C (q) = ⎨?
⎪?
⎩
Part 2: When the next 600,000 units are produced at $100 each:
Total cost = Fixed cost + Cost of first 500,000 units + Cost of remaining units
= 21,600,000 + (115)(500,000) + (100)(q – 500,000)
Note that we multiply 100 by (q – 500,000), because we’re only interested in the
number of units produced after the first 500,000 are already accounted for.
Chapter 2 – Page 32
Next, we’ll use some algebra to simplify the expression:
=21,600,000 + (115 )( 500,000 ) + (100 )( q – 500,000 )
= 21,600,000 + 57,500,000 + 100q − 50,000,000
= 100q + 29,100,000
This gives us the next piece of our Cost Function:
⎧115q + 21,600,000 if q ≤ 500,000
⎪
C (q) = ⎨100q + 29,100,000 if 500,000 < q ≤ 1,100,000
⎪?
⎩
Part 3: When more than 1,100,000 units are produced at $90 each:
Total Cost =
Fixed cost + Cost of first 500K units + Cost of the next 600K units + Cost of remaining
units
= 21,600,000 + (115)(500,000) + (100)(600,000)+90(q – 1,100,000)
Note that we multiply 90 by (q – 1,100,000,000), because we’re only interested in the
number of units produced after the first 1,100,000 are already accounted for.
Next, we’ll use some algebra to simplify the expression:
= 21,600,000+ (115 )( 500,000 ) + (100 )( 600,000 ) + 90 ( q – 1,100,000 )
= 21,600,000 + 57,500,000 + 60,000,000 + 90q − 99,000,000
= 90q + 40,100,000
Answer: This gives us the next piece of our Cost Function:
⎧115q + 21,600,000 if q ≤ 500,000
⎪
C (q) = ⎨100q + 29,100,000 if 500,000 < q ≤ 1,100,000
⎪90q + 40,100,000 if q > 1,100,000
⎩
We can now graph these three functions together to see what our Cost Function looks like
visually. We will use Graphmatica to do that. We enter each equation with the appropriate
domain modifies as follows:
y = 115x+21600000 {0,500000}
y = 100x+29100000 {500000,1100000}
y = 90x+40100000 {1100000,}
However, these graphs are not going to automatically show up in your window, since the default
setting is from -10 to +10 in each direction. In this case, we have values of x that go up to and
Chapter 2 – Page 33
beyond 1,100,000, and we have values of y that are in the tens of millions. To view the graph, we
need to change the grid range appropriately. To do this, we use the “Grid Range…” command in
the View Menu:
This command allows us to change the left/right and bottom/top values that we want to see on the
graph. In this case, left/right values that make sense are 0 and 1,500,000 (0 units up to 1,500,000
produced). Bottom/top values that make sense are 0 to 200,000,000 ($0 up to about
$200,000,000). We can enter these, re-graph, and then adjust as necessary. Here’s what we get
with these values:
Chapter 2 – Page 34
Notice that we can’t see the axes sine we are starting at (0,0) and that Graphmatica displays
values on the axes in scientific notation. 10^6 represents 1,000,000, 10^7 represents 10,000,000,
and 10^8 represents 100,000,000. There’s nothing we can do about the scientific notation since
it’s the way the software works. But we can zoom out once to see the axes:
Chapter 2 – Page 35
Chapter 2 – Page 36
We can then box the area we want to focus on (to eliminate extra blank space) to get the
following:
Keep in mind that y represents total dollars, and x represents total units produced. If we want to
change the labels to better reflect that, we can do so in the Options menu, using the “Graph
Paper…” command:
Chapter 2 – Page 37
Once there, we can select the “Labels” tab:
Chapter 2 – Page 38
Notice that I’ve given the graph a title, “Cost Function,” and renamed the Axis Labels as “Units”
(instead of “x”) and “Cost ($)” (instead of “y”). When OK is clicked, we get the following nice
graph:
Chapter 2 – Page 39
As you know, there are really three pieces to this graph. They are a bit hard to see because each
linear piece has a slope very similar to the others. But if we remember that the cut-off points are
at 500,000 units and 1,100,000 units, it’s easy to identify them:
Up to 1,100,000
units produced.
Up to 500,000 units
produced.
Chapter 2 – Page 40
Chapter 2 Practice Problems 1)
Suppose you have [a,b] = [3,20] and n = 15 intervals. What is the value of Δx?
2)
Suppose you have [a,b] = [−5,35] and n = 12 intervals. What is the value of Δx?
3)
Suppose you have [a,b] = [−10,10] and n = 200 intervals. What is the value of Δx?
4)
Suppose you have [a,b] = [7,40] and n = 250 intervals. What is the value of Δx?
5)
Suppose you have [a,b] = [4,15] and n = 150 intervals. What is the value of Δx?
6)
Suppose you have [a,b] = [−20,−8] and n = 150 intervals. What is the value of Δx?
7)
Suppose you have [a,b] = [2.5,18.8] and n = 60 intervals. What is the value of Δx?
8)
Suppose you have [a,b] = [−8,10] and n = 300 intervals. What is the value of Δx?
9)
Suppose you wanted to graph a function on the interval [-2,2]. Suppose we want to use 401
evenly spaced points. Fill in the first values of x in the table below.
x
-2
10) Suppose you wanted to graph a function on the interval [-2,2]. Suppose we want to use 601
evenly spaced points. Fill in the first values of x in the table below.
x
-2
For Problems (11) to (16), assume that a numerical value for the variable x is in cell A1.
11) Give the Excel command that would compute f ( x) = 3 x + 2 in a different cell.
12) Give the Excel command that would compute f ( x) = πx −
Chapter 2 – Page 41
π
2
in a different cell.
13) Give the Excel command that would compute C (q ) = −.005q 2 + 1.3q − 3.5 in a different
cell.
14) Give the Excel command that would compute g ( x) =
1
in a different cell.
x+5
15) Give the Excel command that would compute P (t ) = 2 t in a different cell.
16) Give the Excel command that would compute h( x) = 5.2e −0.5 x in a different cell.
For Problems (17) to (20), set up an Excel spreadsheet from scratch to graph the indicated
function over the interval [−10,10] using 80 subintervals. (Do not use Graphing.xls or any other
graphing tools to do these problems.) For each problem, copy and paste only the first few rows of
your x−y table as well as a neat, formatted graph as your solution.
17)
f ( x) =
1
(x + 15)
18)
f ( x) =
3
x−7
19)
g ( x) = −.0034 x 2 + 0.06 x + 28.89
20)
g ( x) = .04 x 3 + 2.3 x + 4
⎧2 x + 3 x < 3
⎪
21) Sketch the following piecewise function by hand: f ( x) = ⎨ 1
21
⎪⎩− 2 x + 2 x ≥ 3
⎧x 2 − 4x + 7 x < 4
22) Sketch the following piecewise function by hand: f ( x) = ⎨
⎩− 2 x + 15 x ≥ 4
For Problems (23) to(27), use Graphmatica or Excel to graph the piecewise function.
23)
⎧− x + 2 if x < 2
h( x ) = ⎨
2
⎩− ( x − 4) + 4 if x ≥ 2
24)
⎧4 x 2 − 0.8 x if x < 0
h( x ) = ⎨ 2
⎩4 x + 15 x if x ≥ 0
Chapter 2 – Page 42
25)
⎧5 + 1.5x if x < 0
⎪
T ( x) = ⎨5 - x if 0 ≤ x < 5
⎪2x - 10 if x ≥ 5
⎩
26)
⎧6 + 3x if x < −2
⎪
T ( x) = ⎨ x 3 if − 2 ≤ x < 6
⎪4 if x ≥ 6
⎩
27) Graph the following Cost function over the interval [0,30000].
⎧40000 + 2.5 x if x < 10000
⎪
C ( x) = ⎨45000 + 2 x if 10000 ≤ x ≤ 20000
⎪75000 + 0.5 x if x > 20000
⎩
In words, explain what this Cost function means to someone who may be thinking about
buying some of these items.
For Problems (28) to (34) Use Graphmatica to graph the following functions in the interval
[−10,10].
28)
f ( x) =
1
(x + 15)
29)
f ( x) =
3
x−7
30)
g ( x) = −.0034 x 2 + 0.06 x + 28.89
31)
g ( x) = .04 x 3 + 2.3 x + 4
32)
f ( x) = 4.1e 2.3 x
33)
f ( x) = 2.348 x 4 − 5.287 x 3 − 8 x 2 + 1.88 x − 4.56 .
34) Graph the function N ( x) =
35)
a.
b.
c.
d.
e − 0 .5 x
2
2π
on the interval [−3.5,3.5].
Suppose a Cost function is given by C (q ) = 2q 2 + 3600 . Fix q to be 10 units.
Find C (10)
Find C (q + 1) . (Recall that q = 10)
Find C (q + 0.1) .
Find C (q − 0.01) .
Chapter 2 – Page 43
Answers to Selected Problems 1)
1.133
2)
3)
0.1
4)
5)
0.07333
6)
7)
0.272
8)
9)
x
-2
10)
11) =3*A1+2
12)
13) =-.005* A1^2+1.3* A1-3.5
14)
15) =2^( A1)
16)
Chapter 2 – Page 44
17)
y
4
2
-1 0
0
1 0
-2
18)
19)
30 y
20
10
x
-50
0
50
100
20)
21)
10 y
5
x
-10
-5
0
5
-5
-10
22)
Chapter 2 – Page 45
10
23)
10 y
5
x
-10
-5
0
5
10
5
10
-5
-10
24)
25)
10 y
5
x
-10
-5
0
-5
-10
26)
27)
C(x)
10^5
5x10^4
0
10^4
2x10^4
28)
Chapter 2 – Page 46
3x10^4
x
29)
30)
31)
32)
Chapter 2 – Page 47
33)
34)
35)
a.
b.
c.
d.
3800
3842
3804. 02
3799.6002.
Chapter 2 – Page 48
Chapter 3: Trendlines & Demand Functions Reading Questions 1.) Describe what Trendlines are in your own words and describe how they might be useful.
You should be able to do this in one or two sentences, at the most.
2.) Why are units so important in the Project? Go back to Chapter 1 and summarize what the
units are for each of the important Project variables/quantities.
3.) What is a Demand function? What does it tell you? Be specific. You should be able to do
this in one or two sentences, at the most.
4.) Suppose D(q) = −1.2q + 78 . What is the value of D(20) and, in your own words, describe
what the result means. Show your calculations.
5.) Suppose D(q) = −1.2q + 78 . If the price of each item is $20.4, how many items can you
expect to sell at that price? Show your calculations and explain the result in words.
6.) Do Problem #1 from the Practice-Problems section of this Chapter. Report the Trendlines
and the final result.
7.) Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 3.1: Introduction We begin with a definition:
Definition: Trendline
A trendline is a smooth curve or function that best fits a given set of raw data. It is also called
a "best−fit curve" or a "least squares regression line."
Basically, trendlines are formulas that have been extracted from a set of data. We use trendlines
when we have a series of data in ( x, y ) format and we want to find the equation of a line,
quadratic, or some other function that fits the data relatively well.
The first step to creating a trendline is to graph the points using the (XY) scatter plots in Excel's
Chart Wizard. After that, we run the trendline command.3
3
Much of the material in the first part of this chapter is review material from CSC102B. For more
examples of how to construct trendlines, see the 102B Excel tutorial. We will repeat one of the
examples from that tutorial here to refresh our memories, however.
Chapter 3 – Page 49
Section 3.2: Constructing a Linear Trendline Let’s go through creating a trendline in some detail.
Example 13 The file Shrinkage.xls has data about the amount of weekly merchandise lost to shoplifting
and damage (called shrinkage) as it relates to the average number of clerks on duty. A
sample of 7 weeks worth of data is shown.
Managers want to see if there is a relationship between these two variables and if so, they
want to predict shrinkage from the number of clerks on duty.
First, let’s graph this data using a simple (XY) scatter plot without any lines connecting the
data points. As a review, you may want to open the file and make sure you can get a graph
like this:
Chapter 3 – Page 50
Next, RIGHT click on one of the blue data points until the menu appears:
We will choose the “Add Trendline…” option. You should see this screen:
Chapter 3 – Page 51
This is where your intermediate and college algebra background will come in handy. You’ll
see a list of possible functions that could model this data. In this case, our data looks very
linear so we’ll keep the default choice of “Linear.”
Do not click the Close button yet!
We want to display the linear equation that best fits this data, so check the “Display
Equation on chart” box:
This should generate the following graph when you click the Close button:
You will notice that a straight line has been drawn on the graph that corresponds to the
linear formula y = −3.0328 x + 52.508 . This equation is called the “line of best fit” because
it is the line that best fits this data. (No other line comes closer to fitting all of these points.)
Chapter 3 – Page 52
Now, the equation that shows displays only a certain number of decimal places. If you want
more (or less) you can edit that as well. To do so, RIGHT-click on the trendline equation
and choose “Format Trendline Label…” from the menu:
On the window that appears, choose the “Number” option from the Category list and then
type in how many decimal places you want displayed. We’ll do 10 for this example:
Chapter 3 – Page 53
This gives you the equation with many more decimal places showing:
IMPORTANT NOTE: The default number of decimals that show in a trendline (usually
three) are NOT necessarily rounded for you. To make sure that you have the level of
accuracy you desire, you should absolutely tell Excel to display several decimal places (as
we just demonstrated) and then round appropriately!!!
Section 3.3: The LINEST Command Sometimes, you may want the actual numbers that appear in the trendline equation to be
deposited into a cell so you can compute with them. We will find this very useful when doing
calculations with our project data. To do this, we use the LINEST command, which has the form:
LINEST(known_y’s, [known_x’s],[const],[stat])
For now, only the known-y’s and known_x’s are useful for us. They are, of course, just the y and
x values, respectfully, used to create the graph. To use this command, you highlight TWO empty
cells…one for the slope of the line and one for the y-intercept. We’ll do this in cells E2 and F2.
Chapter 3 – Page 54
Like the FREQUENCY command, we start typing the command directly but don’t use the
ENTER key right away. The command looks like this for this example:
Since we want Excel to compute for more than one cell, we have to use SHIFT-CTRL-ENTER so
that Excel computes values for both cells. When we do this we get the following:
We can now increase the size of these cells and tell Excel to show more decimal places.
More importantly, we can compute with these numbers without losing any accuracy through
rounding.
We can also use the LINEST command with quadratic functions by revising the command just a
little. The following example illustrates this.
Example 14 Use the LINEST command on the following data to create a quadratic trendline.
Chapter 3 – Page 55
Solution:
We start by creating a scatter plot graph of this data:
We can see the data is curved slightly, so a quadratic may fit it well.
In order to add a quadratic trendline, we choose Polynomial from the “Format
Trendline” box as below. We make sure it’s Order is 2, since that’s the degree (order)
of a quadratic.
Chapter 3 – Page 56
This produces the following picture and trendline:
To get the same coefficients using the LINEST command we, select three cells (one for
each coefficient) in a row and type the following command:
Note how the command has changed a bit. The actual command is this:
= LINEST(B2:B8,A2:A8^{1,2})
This follows the general format of:
=LINEST(Known Y’s,Known X’s^{1,2})
It’s important to notice that after the known x values, the command requires “^{1,2}”.
Basically, this tells Excel, that you want to compute coefficients for x that include the
first and the second power. If you remember to use SHIFT-CTRL-ENTER, Excel gives
these coefficients in those cells:
Chapter 3 – Page 57
As before, these numbers retain a lot more accuracy than reading the numbers off the
trendline that Excel displays on the graph. In fact, simply by expanding the size of the
columns, we can see more decimal places of accuracy:
With accurate values like these, we can use them to evaluate this quadratic at any value
of x that is reasonable. For example, we can check the y value when x = 2.5:
We can get this with the following formulas. You should study them to make sure you
see how they work and why they make sense.
Chapter 3 – Page 58
Section 3.4: A Review of Function Shapes Linear and quadratic trendlines are certainly not the only kind that we can find. It may be that the
data we have is modeled better by a quadratic or an exponential function. For this reason, it's
important to have an idea of what various functions basically look like when graphed.
Here is a table that presents this important information:
Name
Linear
Equation
y = mx + b
Shape
y
Example:
y = 3x + 2
x
Quadratic
y = ax 2 + bx + c
y
Examples:
y = x2 + 5x + 4
y = − x 2 + 4 x + 21
Recall that a determines if
the parabola opens upwards
or downward. Important
Note: Excel also calls this a
"Polynomial of Order 2."
Chapter 3 – Page 59
x
Name
Equation
Exponential y = aebx
Shape
y
Example:
y = 4e.05 x
x
Logarithmic
y = a log(bx)
y
Example:
y = 6 log(2.2 x)
x
Power
y = ax n ,
n not an integer
y
Example:
y = 3x1.35
They may look like parts
of parabolas but they are
not the same. (It makes
sense they are close… x 2
is very similar to x1.35 .
x
Chapter 3 – Page 60
Mathematical analysis of business situations often requires formulas for the functions that are
used as models. Unfortunately, the real world almost never provides us with formulas. What
we do find in business are data points that provide approximate values for the models. Trendlines
allow us to use data points to generate formulas which can be used for business planning.
Example 15 A county Child Protective Services, CPS, agency has records showing the number of cases that it
has handled during the years from 1990 through 1998.
Year
1990 1991 1992 1993 1994 1995 1996 1997 1998
Case Load 2373 2825 2806 3534 3808 4312 5067 5368 6445
Noting that the case load is increasing rather rapidly, the agency would like to predict how many
cases it is likely to handle in the year 2002.
Solution:
The first step is to get a graph of this data in front of us. Using Excel and the XY scatter
plot option, we get the following preliminary graph:
7,000
6,000
Cases
5,000
4,000
3,000
2,000
1,000
0
0
2
4
6
8
10
Years Since 1990
An important thing to note is that we have converted the variable on the horizontal axis
to years since 1990…this is pretty standard in cases where time is the independent
variable and you should get into the habit of doing this yourself.
Chapter 3 – Page 61
Looking at this graph, we see the data could be fit by a line or perhaps a quadratic, or
even an exponential function.
y = 486.95x + 2112
7,000
Chart I: Linear
6,000
Cases
5,000
4,000
3,000
2,000
1,000
0
0
2
4
6
8
10
Years Since 1990
y = 34.339x2 + 212.24x + 2432.5
7,000
Chart II: Quadratic
6,000
Cases
5,000
4,000
3,000
2,000
1,000
0
0
2
4
6
8
10
Years Since 1990
y = 2373.6e0.1217x
7,000
Chart III: Exponential
6,000
Cases
5,000
4,000
3,000
2,000
1,000
0
0
2
4
6
8
10
Years Since 1990
But which one is best? There is not necessarily a right answer to that question.
However, let's consider the following: (i) The data points appear to increase and curve
slightly upward. (ii) Many changing populations have patterns of exponential growth.
(iii) Experience in public administration around the country suggests that social service
case loads increase exponentially. For these reasons we will assume that the number of
cases handled in a year that is t years after 1990, is given by an exponential function. ■
Chapter 3 – Page 62
Important: The choice of a basic form for a trend line is of major importance. Further training
in business will help you recognize the appropriate type of function to use in common situations.
However, at the minimum you need to be familiar with the basic shapes of the functions pictured
earlier so you at least know what your options are.
Section 3.5: Dealing with Units In this project we will be working with important functions like Cost, Demand, Revenue and
Profit. However, because these functions often have units associated with them that are not
straightforward, we will first try to lay a foundation that will help us organize all of the units in
our project. For example, a Cost function could ask us to input thousands of units of production.
Hence, we would enter in q = 500 for a quantity of 500,000 units. Also, note that companies often
report their earnings or production in units like millions or billions: Coca Cola reported producing
1.2 billion cases of non−carbonated beverages (such as Minute Maid orange juice) in 2002. This
keeps them from having to type 1,200,000,000, which would get very tedious both to the author
and to the reader of the report or document.
In this project, we will be working with what we will call "converted" units. Some examples will
give us an idea how these work. Some are simple, but they get increasingly complex.
Example 16 Convert 45,310,000 cans of soda to millions of cans.
Solution:
To convert, we divide by 1,000,000. This is the same as moving the decimal point 6
places to the left.
Answer: 45.31 million cans of soda. ■
Example 17 Convert 45,310,000 cans of soda to thousands of cans.
Solution:
To convert, we divide by 1,000. This is the same as moving the decimal point 3 places
to the left.
Answer: 45,310 thousand cans of coke. This may be a strange way to express this −
"Forty−five thousand three hundred ten−thousand cans of soda" − but it is the correct
conversion. ■
Chapter 3 – Page 63
Example 18 Assume that total costs are measured in millions. If a financial report states that the total costs for
2001 was $48.9, what does that mean.
Solution:
In this case, we are converting in the opposite direction from the previous two
examples, so now we need to multiply by 1,000,000. This is the same as moving the
decimal point 6 places to the right.
Answer: Total costs were $48,900,000, or $48.9 million.
Section 3.6: Demand Functions At the very beginning of this project, we said that a Demand function provides us with the
relationship between the number of units that a company can sell and the price per unit the
company can charge to the public. In a monopoly situation, it is presumed that if you increase the
price of a product then the quantity demanded will decrease. Conversely, if the price of product is
decreased then more people should be willing to purchase it since it's more affordable.
Example 19 The Demand function for product is given by the equation D( q) = −3q + 500 , where q is the
number of items produced/sold, and D(q ) is price per unit that can be charged at a level of q
units of production.
a.
If the company expects to sell 100 units, what price can they expect to charge for each of
these units?
b.
If the company would like to charge $350 for a unit, then how many units can they expect
to sell?
c.
What does the slope of the Demand function tell us?
Solution:
Part a:
We are given q = 100 and asked to find D(q) , which is the corresponding demand
price. Substituting q = 100 into the demand functions gives:
D(100) = −3(100) + 500
= −300 + 500
= 200
Answer to Part a: They can expect to charge $200 per item when selling 100 units of
this item.
Chapter 3 – Page 64
Part b:
We are given the demand price, D( q) = $350 and asked to find q, the number of units
that can be expected to be sold. Here, we substitute again and solve for q:
350 = −3q + 500
350 − 500 = −3q
− 150 = −3q
− 150 − 3q
=
−3
−3
50 = q
Answer to Part b: The company can expect to sell only 50 units if they charge $350
per item.
Part c:
The slope of the demand function is –3. The units on the demand function are dollars
per unit.
Answer to Part c: The demand price for this item is decreasing by $3 per unit. This
means that for every additional unit they want to sell, they will have to lower the price
by $3. ■
Example 20 Assume that q is in thousands of items, and the Demand function is give by the equation:
D(q ) = −0.0003q 2 − 0.04q + 23.56
where q is in thousands of units, and D(q ) is in dollars.
What would be the selling price if q = 46.5 (thousand) items are to be sold?
Solution:
Because units are in thousands, we know that q = 46.5 means we really have 46,500
units. But, to do our computations, we will use the q = 46.5 value since we are
specifically told that assume that q is thousands of items. We now use the Demand
function to find the price per unit we can expect to receive at this level of production:
D( q) = −0.0003q 2 − 0.04q + 23.56
= −0.0003(46.5) − 0.04(46.5) + 23.56
= 21.05
2
This is consistent with the
assumption that q is in thousands.
Answer: The price per unit to be expected is $21.05 (since the Demand function is said
to be in terms of dollars.)
Chapter 3 – Page 65
Example 21 A regional restaurant chain plans to introduce a new steak dinner. Managers have tested various
prices in their establishments and arrived at the following estimates for weekly sales:
Price
$14.95 $19.95 $24.95 $29.95
Number sold per week 2,800 2,300 1,600
300
For example, they believe that D( 2,800) is about $14.95. That is, they believe they will sell 2,800
of the new steak dinners at $14.95. Their experience also tells them they are dealing with a
quadratic Demand function. What is the quadratic trendline that best fits this data and what does
it predict about price per dinner when the goal is to sell 2,000 dinners per week?
Solution:
Using Excel’s trendline feature, we get the following graph and equation with this data:
If the goal is to sell 2,000 steaks per week, the graph gives us an estimate of about
$22.00 per steak. However, we can use the trendline to check this estimate:
D( 2 ,000 ) = − 0.0000018 ( 2000 ) − 0.0002953 ( 2000 ) + 30.1899153
2
= 22.40
Answer: The trendline predicts that the sale price should be $22.40 if we expect to sell
2,000 dinners per week.
Chapter 3 – Page 66
A LINEST command produces the same coefficients:
Of course, the cells above are showing many more places of accuracy than the graph
above does. This is an advantage of the LINEST command.
■
Chapter 3 – Page 67
Section 3.7: Focus on the Project In this project, we want to know what the Demand function for our product is before we can
proceed. We have been given information from test markets about how many drives we can
expect to sell in different parts of the country. We will take this information and transform it into
a Demand function that is quadratic. But we first need to be clear on what the units are.
1) Prices for individual drives are given in dollars.
2) Projected quantities of drives in the test markets are actual numbers of drives.
3) Projected quantities of drives in the national market are given in thousands of drives.
Here is the data we have been given for the Class Project:
Potential market:
120,000,000
Test Markets
Market Number
Market Size
Price
Projected Yearly
Sales
(# of drives)
1
1,956,000
$119.95
14,563
2
1,044,000
$129.95
7,143
3
492,000
$139.95
3,179
4
1,512,000
$154.95
8,404
5
1,104,000
$169.95
5,297
6
1,224,000
$179.95
4,573
The first thing we want to do is to take our test market data and use it to project how many items
we can sell in the national market at particular prices. Notice that each test market is associated
with a different price point. Hence, when we put all of the data together we will have a picture of
how this new product might sell if we price it to the public at different levels. For each test
market, we need to determine how many hard drives we can sell at the national level based on the
data from test markets. To do so, will use a basic proportion:
Note that this is the number we are seeking
to find for each test market.
Projected test market sales Projected national market sales (thousands)
=
Size of the test market
Size of the potential national market
Chapter 3 – Page 68
Let’s denote the “Projected national market sales (thousands)” as x, our unknown variable, do an
example for one of the test markets so we can get a good idea of how this proportion works. For
Test Market #1, the resulting proportion would be the following:
14,563
x
=
1,956,000 120,000,000
What this proportion says is that the number of drives we can expect to sell in the national market
(x) is related to the number of drives we expect to sell in the test market. Specifically, the ratio of
items sold to potential customers in each case is the same. Solving for x involves multiplying
both sides by 120,000,000 and doing some arithmetic:
14,563
x
=
120,000,000
1,956,000 120,000,000
893,435.58 = x
120,000,000
Hence, at the price of $119.95 per drive, we expect to sell about 893,436 drives nationally. The
one thing to remember here is that drives in the national market are supposed to be in thousands,
so we divide by 1,000 to get 893.4 (thousand) drives.
If we do the same thing for each of our other five test markets, we get the following results:
Market
#
1
Market
Size
1,956,000
Price/
Drive
$119.95
Projected Market Sales
(Number of Drives)
14,563
Projected National Sales
(K's of Drives) 4
893.4356
2
1,044,000
$129.95
7,143
821.0345
3
492,000
$139.95
3,179
775.3659
4
1,512,000
$154.95
8,404
666.9841
5
1,104,000
$169.95
5,297
575.7609
6
1,224,000
$179.95
4,573
448.3333
4
Important Note: These numbers are rounded from the original Excel cells. The trendline computation that
follows does NOT use the rounded numbers but the original values in the Excel sheet. Be cautious when
you work your team data.
Chapter 3 – Page 69
The important data in this table that we need for our Demand function are the columns for
Price/Drive and Projected National Sales (K’s). If we put these numbers into their own table we
get this:
Projected National Sales
(K's of Drives)
893.4356
Price/
Drive
$119.95
821.0345
$129.95
775.3659
$139.95
666.9841
$154.95
575.7609
$169.95
448.3333
$179.95
For the class project, we are told that the Demand function is quadratic so this data gives us a
basis upon which to run a trendline. Doing so in Excel gives us the following results and graph:
Remember, x represents q, quantity sold.
Chapter 3 – Page 70
Notice that the leading coefficient appears to be zero. But it’s not! The issue here is that Excel
may not display more than two trendline decimal places. To get Excel to show more decimal
places, select the equation and then right click on the equation. Choose the “Format Trendline
Label…” command.
Change the Category to “Number” and choose 8 decimal places of accuracy. That will give you
this:
Hence, our trendline5 is:
D(q) = −0.00011519q 2 + 0.01501173q + 197.22734383
5
IMPORTANT NOTE: Unless otherwise specified, we will use the coefficient values that the LINEST
command gives us rather than rounded values. This insures more accuracy in future calculations.
Chapter 3 – Page 71
Of course, the best way to get the same result is to use the LINEST command. Doing so gives the
following coefficients, with several decimal places of accuracy shown:
Example 22 What price can we expect to charge per drive if we want to sell 750,000 drives?
Solution:
We will first need to covert q to thousands, since the demand function requires it.
750,000 is the same as 750 thousand, so q = 750. Substituting that into the demand
function gives the following:
D(750) = −0.00011519 ( 750 ) + 0.01501173 ( 750 ) + 197.22734383
2
= −0.00011519 ( 562,500 ) + 11.2586475 + 197.22734383
≈ −64.7943 + 11.2568 + 197.2273
= 143.68926858
Note: We’ve reduced the number of decimal
points because once we have numbers
substituted, we don’t need to worry about all the
place values. After all, D(q) is in dollars, so we
only need 2 decimal places.
≈ 143.69
Answer: Remember that D(q ) is in thousands, so this result tells us that we can expect
to sell 750,000 drives for about $143.69 each. ■
You are now ready to do similar computations for your team’s data.
Chapter 3 – Page 72
Chapter 3 Practice Problems 1)
a.
b.
c.
The data shown in the table to the right is found in the file
called Total Costs.xls. Use the data to find the following:
A linear trendline that fits this data. Include a graph of
your result.
A quadratic trendline that fits this data. Include a graph of
your result. (Remember, a quadratic is a polynomial of
order 2 in Excel.) Include a graph of your result.
A exponential trendline that fits this data. Include a graph
of your result.
Units
Produced
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
11000
12000
13000
14000
Total
Cost
78055.81
122263.4
253024.3
242359.9
447668
599034.3
675202
436541.3
737986.7
624661.4
885523
694889.3
1011783
1352986
2)
a
b.
Use the data in the file Demand.xls to find the following:
A linear trendline that fits this data. Include a graph of your result.
A quadratic trendline that fits this data. Include a graph of your result. (Remember, a
quadratic is a polynomial of order 2 in Excel.) Include a graph of your result.
3)
During the first 6 days of a special promotion, a small business records the following sales
information. The dollar amount listed for each day is the total cumulative sales from the
start of the promotion.
Day
1
Total Sales $29,390
a.
b.
c.
4)
2
$37,358
3
$47,116
4
$50,604
5
$57,066
6
$60,666
Fit a Power trend line to the data and use it to estimate the total sales during the first 9 days
of the promotion.
Fit a Linear trend line to the data. Include a graph as part of your solution.
Which model do you think is more realistic? Explain your decision. Include a graph as part
of your solution.
Fit an exponential trend line to the following data and use the line’s formula to predict the
highest closing price during July. Include a graph as part of your solution.
Highest Monthly Close of MONEY.com Stock Shares
Month Name
January February March
April
May
Month Number
1
2
3
4
5
Highest Close
$65.35
$65.95
$67.71
$68.62
$71.24
Chapter 3 – Page 73
5)
The following table gives the U.S. population in 10-year increments.
Year
1790
1800
1810
1820
1830
1840
a.
b.
c.
6)
Population
3.9
5.3
7.2
9.6
12.9
17.1
Year
1850
1860
1870
1880
1890
1900
Population
23.1
31.4
38.6
50.2
62.9
76.0
Year
1910
1920
1930
1940
1950
1960
Population
92
105.7
122.8
131.7
150.7
179
Year
1970
1980
1990
2000
Population
205
226.5
248.7
281.4
After converting the Years variable to "Years since 1790," find an exponential model that
fits this data best.
Create a graph of this data and the exponential model.
Use your equation to predict the 2000 US population and compare it to the actual value
reported above. How close is the prediction to the actual number in the table? Explain why
you think this worked out, or did not work out to be close.
Fuller and coworkers (1997) estimated the costs of a medium-sized cotton gin plant as
shown in the following table:
x
y
2
217.6
4
271.12
6
325.5
8
380.8
10
437
12
494.16
15
581.7
20
732
x represents the annual number of bales produced in thousands.
y represents the total cost in thousands of dollars.
a.
b.
c.
Determine the best-fitting linear and quadratic models of this data.
Create a graph of this data and these two models.
Which is better, linear or quadratic? Explain your answer.
7)
TIAA−CREF is the largest retirement program in
Rate of Return First Year Income $
the world. They cater specifically to educational
0%
32,900
and research institutions. Every year, they send
3%
55,500
out “Benefit Reports” which give participants an
6%
98,856
idea of what their future retirement will be if
9%
183,550
certain factors (inflation, contribution levels,
12%
250,700
rates of return) stay fixed. Suppose a participant
receives a Benefit Statement which has the data given in the table. The Rate of Return is
that earned on contributions paid by the participant and First Year Income is the amount of
first-year income the participant would receive (in 30 years) on that rate of return, all other
things being equal.
a.
b.
c.
Use Excel and Trendlines to come up with an exponential equation that models this data.
Graph the data.
What does this model predict for the First-Year Income with a rate of return of 10.5%?
Show your calculations.
Chapter 3 – Page 74
8)
A tourist company charged $60 per person for a sightseeing trip and got 40 people to sign
up for the trip. Past experience and data tells them that for the same trip, each $2 increase in
the price above $60 results in the loss of one customer. The company has fixed costs of
$300 per trip and variable costs of $4 per person. The following data is found in the file
called Tourist.xls and give information on the revenue, costs, profits, etc. for this scenario.
People Price/Person Revenue
40
60
2400
39
62
2418
38
64
2432
37
66
2442
36
68
2448
35
70
2450
34
72
2448
33
74
2442
32
76
2432
31
78
2418
30
80
2400
29
82
2378
28
84
2352
Etc..
Cost
460
456
452
448
444
440
436
432
428
424
420
416
412
Profit
1940
1962
1980
1994
2004
2010
2012
2010
2004
1994
1980
1962
1940
a.
b.
c.
d.
Find an appropriate trendline that gives the Revenue as a function of People on the trip.
Find an appropriate trendline that gives the Cost as a function of People on the trip.
Find an appropriate trendline that gives the Profit as a function of People on the trip.
Produce a graph of the Profit function. For how many People on the trip are Profits
maximal?
9)
a.
b.
c.
See the description and data as described in Problem (8). Do the following:
Find an appropriate trendline that gives the Revenue as a function of Price/Person.
Find an appropriate trendline that gives the Cost as a function of Price/Person.
Find an appropriate trendline that gives the Profit as a function of Price/Person.
10) The Demand function for a product is given by D(q ) = 150 − 6q 2 . What is the value of
D(2) and explain what it means. Assume D(q ) is in single dollars and q is in single units.
11) The Demand function for a product is given by D(q ) = 850 − 4q 2 . What is the value of
D(12) and explain what it means. Assume D(q ) is in single dollars and q is in single units.
12) The Demand function for a product is given by D(q ) = 156 − 0.25q 2 . What is the value of
D (20) and explain what it means. Assume D(q ) is in single dollars and q is in single units.
Chapter 3 – Page 75
1
13) Suppose the demand for a certain textbook is given by: D(q ) = − q 2 + 40 dollars, where q
5
is the number of thousands of textbooks. How many textbook are in demand…
a.
At a price of $10?
b.
At a price of $20?
c.
At a price of $30?
d.
At a price of $40?
14) Graph the Demand function from Problem (13) and verify that your answers make sense.
15) In Managerial Economics, the author (Spencer) gives a Demand curve for beef. The data
are given in the following table:
Year 1947 1948 1949 1950 1951
.0358 .0390 .0335 .0345 .0385
x
69.6 63.1 63.9 63.4 56.1
y
Year 1952 1953 1954 1955 1956
.0356 .0261 .0261 .0243 .0224
x
62.2 77.6 80.1 82.0 85.4
y
Year 1957 1958 1959 1960
.0226 .0253 .0246 .0228
x
84.6 80.5 81.6 85.8
y
x is the price of beef divided by disposable income per capita; y is the beef consumption per
capita in pounds.
a.
b.
Find a linear trendline for this data and provide a graph of the trendline.
Does the line slope down? What does this say about the price of beef and the demand for
beef?
Chapter 3 – Page 76
Answers to Selected Problems 1)
a.
b.
c.
2)
3)
a.
b.
c.
4)
5)
a.
b.
c.
6)
a.
b.
c.
7)
a.
b.
y = 78.903x - 8774.6
y = 0.0018x2 + 52.455x + 61753
y = 121853e0.0002x
y = 29044x0.4125
y = 6256.9x + 25134
y = 63.562e0.0212x
6.0566e0.0202x
The 2000 population is 421.217 million. It’s not very close.
The linear is y = 28.545x + 155.24 and the quadratic is y = 0.115x2 + 26.046x + 165.07.
y = 33677e17.526x
y
10^7
10^6
x
0
c.
8)
9)
10)
11)
12)
13)
a.
b.
c.
d.
14)
15)
a.
b.
0.1
0.2
0.3
Income of $212,097.
D(12) = 274, which means if you sell 12 units you can expect to charge $274 per unit.
About 12,247 books.
About 10,000 books.
About 7,071 books.
0 books.
− 1582.x + 120.5
An increase in price leads to a decrease in demand
Chapter 3 – Page 77
Chapter 4: Revenue and Profit Functions Reading Questions 1.) What is a Revenue function? What does it tell you? Be specific.
2.) Suppose a Demand function is given by D(q ) = −2.2q + 350 .
a.
What is the Revenue function, R (q ) ? Simplify your answer.
b.
How much revenue can you expect to receive by selling 120 units? Show work and
calculations.
3.) Suppose a Revenue function is given by R (q ) = −3.1q 2 + 450q . If the cost function is
given by C (q ) = 5000 + 14.5q
a.
What is the Profit function, P (q )
b.
How much profit can be expected from selling 100 units?
4.) In your own words, explain why it makes sense that R (q) = q × D(q) .
5.) Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 4.1: The Revenue Function So far, we’ve managed to talk about our Total Cost function, C (q ) , and our Demand function,
D(q ) . In this chapter we will look at the two remaining functions that we will need for our
project. The Revenue function, R(q ) , and the Profit function, P(q ) , are the last pieces of our
puzzle. It turns out that once we know the Demand function, we can easily get the Revenue
function and once we know the Revenue and Cost functions, we can easily get the Profit function.
So this chapter is dedicated to completing our library of project functions.
The Revenue function, sometimes called the Total Revenue function, tells us how much money
we can expect to bring into our company by selling a specific number of units of a product. This
is revenue before we take into account any of the costs we incur to produce the item. It’s the
money that we receive from customers who buy the product.
A basic principle that we can use to determine revenue is “Revenue equals total units sold times
the price per unit.” For example, if we sell 100 units for $500 each, then our total revenue will be
100×500 = $50,000. This basic principle can be combined with what we’ve already learned to
help us easily find Revenue functions. Notice this:
Total Revenue = (Units Sold) × (Price per Unit)
The revenue function
should depend on q.
This is just q,
quantity sold.
Chapter 4 – Page 78
Price per unit is given by the
Demand function, D(q ) .
From what we see above, the following definition and equation should now be clear:
Definition: Revenue Function
The revenue function gives the relationship between total units sold and total money (revenue)
brought in from sales of those units.
In words:
Total Revenue = (Units Sold) × (Price per Unit)
In function notation:
R(q) = q ⋅ D(q)
Example 23 The Revenue function for a new piece of technology is given by R (q) = −4q 3 + 8q 2 + 3q , where
q is thousands of items sold and R(q) is millions of dollars.
a.
b.
How much revenue will be received by selling 1,200 units of this new technology?
The maximum revenue that can be hoped for with this new technology is about 9 million
dollars. How many units will need to be sold in order to achieve this maximum revenue?
Solution:
Part a:
Since q is in thousands, we first need to covert 1,200 to thousands by dividing by 1000;
this gives us q = 1.2 (thousand). We now substitute this into our Revenue function:
R (1.2) = −4(1.2 ) + 8(1.2 ) + 3(1.2 )
3
2
= −4(1.728) + 8(1.44 ) + 3.6
= −6.192 + 11.52 + 3.6
= 8.208
Answer to Part a: If 1,200 units are sold, total revenue will be 8.208 million dollars
($8,208,000).
Chapter 4 – Page 79
R(q)
Part b:
There are two ways to answer this question. The first is to graph the Revenue function
and see what it looks like. Using Excel, we get the following result:
10
8
6
4
2
q
0
-2
0
0.5
1
1.5
2
2.5
3
-4
-6
From the graph, we can see that the revenue reaches a maximum when q is about 1.5
(thousand) units. However, this is just a guess. If we wanted a more “exact” answer, or
wanted to verify our guess, we would need to solve our Revenue function to find out
when R(q ) = 9 (million). This would give us the following equation with which to
work:
9 = −4q 3 + 8q 2 + 3q
Unfortunately, we have no easy way to solve this equation algebraically, so we are
stuck with the graphical approach. Later in this project, we will learn about a tool that
Excel has which will help us solve these equations relatively easily, and without
algebra. ■
Example 24 In a previous chapter, recall the problem where a regional restaurant chain plans to introduce a
new steak dinner. Managers tested various prices in their establishments and arrived at the
following estimates for weekly sales:
Price
$14.95 $19.95 $24.95 $29.95
Number sold 2,800 2,300 1,600 300
per week
a.
b.
c.
What is the Revenue function?
How much revenue can they expect to receive by selling 2,100 steaks per week?
Graph the Revenue function and use it to predict what price the managers should set the
price to maximize their weekly revenue.
Chapter 4 – Page 80
Solution:
Part a:
Recall from the previous chapter that the demand function for this data was found to be
D(q) = -0.0000018q 2 - 0.0002953q + 30.1899153
The Revenue function can be found by multiplying Demand by q:
R(q) = q ⋅ D(q)
(
= q − 0.0000018q 2 − 0.0002953q + 30.1899153
3
)
2
= −0.0000018q − 0.0002953q + 30.1899153q
Answer to Part a: The Revenue function is
R(q ) = −0.0000018q 3 − 0.0002953q 2 + 30.1899153q .
Part b:
To find the total revenue from selling 2,100 dinners per week, we substitute 2,100 into
the Revenue function:
R (2100) = −0.0000018(2100) − 0.0002953(2100 ) + 30.1899153(2100 )
3
2
= 45,426.75
Answer to Part b: Selling 2,100 dinners per week results in total revenue of
$45,426.75. ■
Part c:
The graph of the revenue function is easily generated in Excel using methods we’ve
previously discussed. The graph looks like this (generated from Graphmatica):
6x10^4
R(q)
4x10^4
2x10^4
q
0
2000
4000
-2x10^4
Answer to Part c: From this graph, it appears that maximum revenue is achieved by
selling about 2,300 units. The total revenue generated at that level of sales is about
$46,000 per week. We can also check the Demand function to verify that the
corresponding price per dinner would be about $19.99. ■
Chapter 4 – Page 81
Example 25 Assume that q is in thousands of items, and the demand function is
D (q ) = −0.003q 2 − 0.04q + 23.56 ,
where q is in thousands of units, and D(q ) is in dollars. In a previous chapter, we used this same
Demand function to find that if q = 46.5 (thousand) items are sold, then the price per unit would
be $15.21.
a.
What is the total revenue in this situation, given in millions of dollars?
b.
Write the Revenue function in terms of q and simplify it as much as possible. The function
should be in terms of millions of dollars, with q still in thousands.
Solution
Part a:
Total Revenue received is equal to the (price per unit)×(number of units produced/sold).
The price per unit is simply the demand equation, D(q ) , and the number of units
produced/sold is just q. Notice that we already know that we can expect $15.21 per
single unit. Hence we need to make sure we multiply that by the appropriate value of q,
which is 46,500 single units. Hence, our total revenue is:
(15.21)(46,500) = 707,265 which gives us $707,265 of total revenue.
$/unit
×
units
=
$
Because we want revenue in millions of dollars, we need to divide our single
dollars by 1,000,000 to get $0.707265 million.
Answer to (a): Total revenue is about $0.707 million.
Part b:
Total Revenue received is equal to the (price per unit)×(number of units produced/sold).
The (price per unit) is simply the demand equation, D(q) , and the (number of units
produced/sold) is just q. So, our demand function is generally:
R(q) = D(q) ⋅ q
= q ⋅ D(q)
However, this does not take into account our two different unit conversions. If you look
back at Part a, when we converted to revenue, we had to first multiply q = 46.5 by 1000
to get to single units so that we could compute total revenue. That gave us an answer in
regular dollars that we then had to divide by a million since we were asked to compute
revenue.
Chapter 4 – Page 82
So if we do that here to the general equation given just above, we get this:
R (q) = q × 1000 × D(q ) ×
Converts q from
thousands to
single units.
1
1,000,000
Converts R(q) to
millions.
If we clean this up, we get the following general expression for Revenue when q is
given in thousands and R(q ) is given in millions:
q ⋅ D (q) × 1,000
1,000,000
q ⋅ D(q)
R(q) =
1,000
R(q) =
Important!
Finally, we are ready to write our Revenue function:
R(q) =
q ⋅ D(q)
1,000
q ⋅ (− 0.003q 2 − 0.04q + 23.56 )
=
1,000
=
− 0.003q 3 − 0.04q 2 + 23.56q
1,000
Note that in this last step, each term of
the numerator is divided by 1000,
which shifts each decimal point three
places to the left.
= −0.000003q 3 − 0.00004q 2 + 0.02356q
Answer to part (b). The Revenue function is
R (q) = −0.000003q 3 − 0.00004q 2 + 0.02356q . ■
This last example is an important one! Make sure you go back and read it carefully because you
will need to do the same kind of thing for the project.
Example 26 Use the previous example to find the Revenue function for the Class Project, given that we
already know that D( q) = −0.00011519q 2 + 0.01501173q + 197.22734383 .
Solution:
The units from Example 3 exactly match those of our project. Hence, the process
needed to get our Project Revenue function is precisely what we did in Example 3.
Namely:
q ⋅ D(q)
R(q) =
1,000
Chapter 4 – Page 83
So, we have this:
R(q) =
q ⋅ D(q )
1,000
q (− 0.00011519q 2 + 0.01501173q + 197.22734383)
1000
3
− 0.00011519q + 0.01501173q 2 + 197.22734383q
=
1000
= −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
=
Answer: The Revenue function for the class project is:
R (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q . ■
Section 4.2: Profit Functions The final function we need to deal with is the almighty Profit function. This is the bottom line.
Are we making money, or are we losing it? The Profit function will take into account not only
how much money we have received from selling goods to customers but will also factor in how
much money we’ve spent to produce, market, and sell the units.
Definition: The Profit Function
Let P(q ) be the profit function.
Let R(q ) be the Revenue function.
Let C (q ) be the Total Cost function.
Profit = Revenue – Costs
P(q) = R(q) − C (q)
Example 27 Suppose the Cost function for a new product is C (q ) = 5000 + 2q and the Revenue function is
R (q) = 15q (they will make $15 per item), where q is units produced and all functions are in
dollars. What is the Profit function and how much profit do they make by selling 10,000 units?
Solution:
The Profit function is:
P(q) = R(q) − C (q)
= 15q − (5000 + 2q )
= 13q − 5000
Chapter 4 – Page 84
To find profits for 10,000 units, we compute the following:
P (10000) = 13(10000 ) − 5000
= 125,000
Answer: Selling 10,000 units results in $125,000 in profits. ■
Example 28 Suppose you own a bakery, and it’s the only one in town. You’re selling small cakes which have
fixed costs of $200 per week and it costs $0.30 to make each cake. After analyzing data from the
market, you find that the Demand function is approximately: D( q) = −0.000001q 2 − 0.00002q + 4 .
a.
What is the Cost function?
b.
What is the Revenue function?
c.
What is the Profit function?
d.
Estimate the profits, if any, from selling 200 cakes in a week. (Assume you sell all the cakes
you make.)
e.
Graph all three functions on one set of axes.
Solution:
Part a:
The Cost function has both fixed costs ($200) and variable costs ($0.30) so we can
write it quickly:
C (q ) = VC ( q) + C 0
= 0.3q + 200
Answer a: C (q) = 0.3q + 200 .
Part b:
R(q) = q ⋅ D(q)
= q (− 0.000001q 2 − 0.00002q + 4)
= −0.000001q 3 − 0.00002q 2 + 4q
Answer b: R (q) = −0.000001q 3 − 0.00002q 2 + 4q .
Part c:
P(q) = R(q) − C (q)
= −0.000001q 3 − 0.00002q 2 + 4q − (0.3q + 200 )
= −0.000001q 3 − 0.00002q 2 + 4q − 0.3q − 200
= −0.000001q 3 − 0.00002q 2 + 3.7 q − 200
Answer c: P (q ) = −0.000001q 3 − 0.00002q 2 + 3.7 q − 200 .
Chapter 4 – Page 85
Part d:
To find the total profits from selling 200 cakes in a week, we substitute q = 200 into the
Profit function:
P (200) = −0.000001(200 ) − 0.00002(200 ) + 3.7(200) − 200
= 531.20
3
2
Answer d: The profit from selling 200 cakes is about $531.20. That’s sweet!
Part e:
Graphing all three of these graphs on one set of axes is a task we have not done before.
Open the file called Chapter4.xls and look at the sheet called Cakes. There you will see
how we have arranged the data to be graphed. To graph the data, select the cell range
A3:E26 and run the Chart Wizard. Excel will automatically take the first column, q, to
be the horizontal axis, and each of the next four rows will be a separate set of points
graphed independently, as they should be. Here is what happens when you graph with
Excel:
4000
Demand
Revenue
3000
Cost
Profit
2000
1000
0
0
500
1000
1500
-1000
-2000
-3000
-4000
Chapter 4 – Page 86
2000
2500
Using Graphmatica, a slightly smoother picture can be generated:
$
4000
R(q)
P(q)
2000
C(q)
q
0
500
1000
1500
2000
-2000
You can see from either graph that selling 200 cakes is not even close to the number
needed to maximize profits. It appears that the bakery would need to sell closer to 1100
cakes per week to reach this peak. You better buy more ovens! ■
Example 29 Recall the example where a regional restaurant chain plans to introduce a new steak dinner.
Managers estimate that the new menu item will have to support $9,000 out of the complete
weekly fixed cost of operating the chain. Hence, C0 = $9,000. The head chef makes the
following estimates for the costs of preparing various numbers of dinners.
Number of dinners, q
Variable Costs
1000
$14,000
2000
$22,000
3000
$28,000
Notice that there is some economy of scale. The cost of preparing the first thousand dinners is
$14,000, of preparing the second thousand dinners is an additional $8,000, and of preparing the
third thousand dinners is an additional $6,000.
a.
In order to develop a formula for the variable cost function, we will have Excel fit a trend
line through the three estimated data points. The managers know that, in former menu
offerings, food preparation costs have usually turned out to follow power function models.
That is, VC(q) = u×qv, for some constants u and v. Find the appropriate cost function.
b.
From the previous section, we know that the Revenue function for this situation was
R ( q ) = − 0 .0000018 q 3 − 0 .0002953 q 2 + 30 .1899153 q . Use this and other
information we now have to find the Profit function and determine where Profits are
maximized. ■
Chapter 4 – Page 87
Solution:
Part a:
Answer a: Selecting the Power trendline in Excel give us C(q) = 177.03q0.6332.
Part b
To find the profit function, we subtract Costs from Revenue:
Answer b:
P(q) = R(q) − C (q)
= −0.0000018q 3 − 0.0002953q 2 + 30.1899153q − 177.03q 0.6332
This cannot be simplified any further, so we graph it as it is.
P(q)
4x10^4
2x10^4
q
0
1000
2000
3000
400
-2x10^4
From the graph, it appears that profits are maximized at about 2,000 units. (The actual
value is 2,024 units.) ■
Chapter 4 – Page 88
Section 4.3: Focus on the Project At this point, we have developed the three main functions that we need for our Class Project:
Revenue, Costs, and Profit. It's time to put them all together on one graph and see what we are
faced with when making our decisions.
First, however, we need to make an adjustment to the Cost function that we found back in
Chapter 2. At that time, we had not yet paid any attention to units and so our Cost function will
need to be adjusted so it uses the same units that our Revenue and Profit functions use. This will
also assure that all of our units conform to those specified in the Project.
Example 30 Adjust the class project Cost function so it has units that match the Revenue and Profit functions.
Solution:
Given the following data, we first covert the table so that costs are in millions of
dollars, q is given in thousands of drives, and costs per drive are in millions of dollars
per thousands of drives.
Fixed Costs = $21,600,000
Quantity Produced/Sold
Cost Per Drive
First 500,000
$115
Next 600,000
$100
All others beyond 1,100,000
$90
We start by dividing the q values by 1,000 since q values are in thousands.
500,000 → 500
600,000 → 600
1,100,000 → 1,100
We also divide the fixed costs by 1,000,000 since costs are in millions.
21,600,000 → 21.6
The last thing to take care of is the Cost Per Drive column. Since costs are in millions
and drives/units (q) are in thousands, we need to end up with millions of dollars per
1000 drives. To see what happens here, let's look a specific case.
Chapter 4 – Page 89
If we know that in the first batch it will cost us $115 per unit, then we can think of the
conversion as follows:
$115
1000 drives
" unit" of a million dollars
×
×
drive 1 thousand drives
$1,000,000
115 × 1,000 millions of dollars
=
1,000,000 1 thousand drives
The Bottom Line:
115 millions of dollars
=
Divide by 1000!
1000 1 thousand drives
millions of dollars
= 0.115
1 thousand drives
Hence, what we have is that it takes 0.115 million dollars to produce one thousand
drives. That is the same as saying $115 per single drive! Basically, then, we need to
divide all the values in the Cost Per Drive column by 1,0006. Here is our new, adjusted
table, with units converted nicely.
Fixed Costs = $21.6 (millions)
Quantity Produced/Sold Cost Per Drive (millions per thousand drives)
First 500
$0.115
Next 600
$0.100
All others beyond 1,100
$0.90
This table will be used in the next example. ■
Example 31 Use the converted table above to find the cost to produce 1,200,000 drives.
Solution:
First convert 1,200,000 drives to thousands to get q = 1,200 (thousand) drives.
Next, break this up into batches: 1,200 = 500 + 600 + 100
The first
500,000.
The next
600,000.
The last
100,000.
The first 500 (thousand) drives will cost 0.115 million dollars per thousand drives.
500×0.115 = 57.5 million dollars
(Note that the cost is in millions because we carefully constructed our table to
produce a total costs in terms of millions.)
The next 600 (thousand) drives will cost 0.100 million dollars per thousand drives.
600×0.100 = 60 million dollars
6
Be careful about what you assume. We divide by 1,000 here because of the units particular to this project.
For other units or data sets, this may not hold true.
Chapter 4 – Page 90
The last 100 (thousand) drives will cost 0.090 million dollars per thousand drives.
100×0.090 = 9 million dollars
Don't forget that we have 21.6 million dollars in fixed costs!
Therefore, we can add these up: 57.5 + 60 + 9 + 21.6 = 148.1
Answer: 1,200,000 drives will cost $148,100,000. Or, said in terms of our project units,
1,200 (thousand) drives will cost $148.1 (million). ■
Example 32 Write the piecewise functions for the class project’s Cost and Profit functions.
Solution:
For the first 500 (thousand) drives, we know that costs are going up by $0.115 million
dollars per thousand drives (see previous example). We also know we have fixed costs
of $21.6 (million). Hence, we can think of 0.115 as the slope of a line, and 21.6 as the
y−intercept of that line (since it represents the total costs to produce 0 units). Therefore,
the first part of our piecewise Cost function is 0.115q + 21.6 .
⎧0.115q + 21.6 if q ≤ 500
⎪
Answer: Our piecewise cost function begins as: C (q) = ⎨?
,
⎪?
⎩
where q is thousands of drives and C (q ) is millions of dollars. The remaining parts are
left for you to verify (if you choose to do so) but are given below. ■
If you remember, we have already built a piecewise function for the Project's total costs
in a previous chapter. In that chapter's Focus on the Project Section, we found our Cost
function to be the following:
⎧21,600,000 + 115q if q ≤ 500,000
⎪
C o (q ) = ⎨29,100,000 + 100q if 500,000 < q ≤ 1,100,00
⎪40,100,000 + 90q if q > 1,100,000
⎩
Chapter 4 – Page 91
I've changed the name of our previous function to C o (q ) so we don't confuse it with our
newer version. Look at the first entry or line of each function and compare it to the
results below. What do you notice?
Obviously, these two expressions are related. In C o (q ) , if we convert the 21,600,000 to
millions, we get 21.6, and if we convert the 115q to thousands, we get 0.115q. Our
functions would then match each other, which is good news. Notice that the inputs for
each of these two functions are different: C o (q ) requires that we put in actual numbers
of drives (e.g. 450,000) while the new C (q) requires that we enter thousands of drives
(e.g. 450).
From an example in this chapter, we have
R (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
Finally, since P (q) = R (q ) − C (q ) , we also have our Profit function.
Answer:
⎧21.6 + 0.115q if q ≤ 500
⎪
P(q ) = −0.00000011519q + 0.00001501173q + 0.19722734383q − ⎨29.1 + 0.100q if 500 < q ≤ 1,100
⎪40.1 + 0.090q if q > 1,100
⎩
3
2
Putting this all together, we see we have the following "family" of functions:
Project Functions
R (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
⎧21.6 + 0.115q if q ≤ 500
⎪
C ( q) = ⎨29.1 + 0.100q if 500 < q ≤ 1,100
⎪40.1 + 0.090q if q > 1,100
⎩
⎧21.6 + 0.115q if q ≤ 500
⎪
P(q ) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q − ⎨29.1 + 0.100q if 500 < q ≤ 1,100
⎪40.1 + 0.090q if q > 1,100
⎩
Chapter 4 – Page 92
Now, I know what you're thinking! That Profit function is one monstrous and scary entity. Is
there some way we can simplify it? Let's start by graphing Revenue and Cost together:
$ (millions)
200
C(q)
100
R(q)
q (thousands)
0
500
1000
Because P (q ) = R( q) − C (q) , profits will only be positive when R (q ) , Revenue, is greater than
C (q) , Costs. This makes sense. Realistically, we are mainly interested in attaining positive
profits so places where we have non−positive profits will be of little use to us. Graphically, this
corresponds to the subinterval where the graph of R(q ) is above the graph of C (q) . We can
estimate this visually to be values between about 275 and 800 thousand units. Notice that the Cost
function's third piece is valid for values of q greater than 1,110. This is clearly not useful since
profits are negative there, so we can completely omit this third piece of the Cost function when
considering when Profits are positive or maximal. This leads us to the following two possible
Profit functions:
1.)
If q ≤ 500 , then
P1 (q) = R(q ) − C (q)
= −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q − (21.6 + 0.115q )
= −.00000011519q 3 + 0.00001501173q 2 + 0.08222734383q − 21.6
2.)
If 500 < q ≤ 1,100 then
P2 (q) = R(q ) − C (q)
= −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q − ( 29.1 + 0.1q )
= −0.00000011519q 3 + 0.00001501173q 2 + 0.09722734383q − 29.1
Chapter 4 – Page 93
This is a new piecewise function with only two pieces, P1 and P2. If we put them on a graph,
being careful to graph each piece only in the appropriate domain, this is the picture we get:
Notice this little bump in the graph. This point is q = 500, the precise point where
the Profit functions changes its formula…hence, it also changes its graph.
$ (millions)
10
P2
5
P1
q (thousands)
0
200
400
600
800
From the graph shown, we now get a preliminary feel for where our profits are maximized. It
looks as though profits are maximized when between 500 and 600 thousand units are produced
and sold.
Chapter 4 – Page 94
We can put all three on one graph. It looks like this:
$ (millions)
150
C(q)
100
R(q)
50
Bump
q (thousands)
0
200
400
600
800
1000
120
P(Q)
-50
It's hard to see the bump in the Profit graph, but it's there if you zoom in. Also, the graph goes
from 0 to 1,100 to be consistent with the constraints on the Profit function that we saw earlier.
Of course, all of this can be done in Excel with one column on the left for q, and three columns of
data to the right of that for Revenue, Cost, and Profit.
You are now ready to do similar work on your team’s data.
Chapter 4 – Page 95
Chapter 4 Practice Problems 1)
a.
b.
c.
d.
e.
f.
Suppose the Demand function is D(q ) = −1.5q + 300 .
Find the price to charge if 100 units are to be sold.
Find the price to charge if 150 units are to be sold.
If you charge $80 per item, how many units can you expect to sell?
Find the associated Revenue function, R(q ) , and simplify.
What is the Revenue from the sale of 100 units?
What is the Revenue from the sale of 150 units?
2)
Consider the following Demand graph:
p=D(q)
250
200
150
100
50
q
0
a.
b.
3)
a.
b.
c.
d.
e.
100
200
300
400
500
600
Use the graph to estimate the price we can charge per unit if we hope to sell 500 units. In
this case, what would be your total Revenue?
If you were to charge $120 per item, how many units can you expect to sell? In this case,
what would your total Revenue be?
Let the Demand function be given by D( q) = −1.5q + 300 . Suppose fixed cost are
C 0 = 3600 and let the variable costs be $90 per unit.
Find the Cost function, C (q ) .
Find the total Cost to produce 75 units.
Find the total Revenue when selling 75 units. (Hint: See Problem 1d)
Find the total Profit when producing and selling 75 units.
Give the simplified Profit function in terms of q.
Chapter 4 – Page 96
4)
a.
b.
c.
d.
5)
a.
b.
c.
The Alexander Rug Co. produces hand−crocheted rugs. The Cost function is given by
C ( q) = 6,452 + 72q (dollars). They receive $153 per rug when they sell them.
How much does it cost to produce 133 rugs?
Give the correct Revenue function.
Give the expression for the Profit function.
What are their profits when they sell 133 rugs?
Graph the Cost, Revenue and Profit functions all on one set of axes found or described in
the previous problem.
The point where the Revenue and Cost functions are equal called the Break−Even point.
Use the graph to estimate the break−even point.
Set the Revenue function equal to the Cost function and then solve for q to get the exact
break−even point.
What is the value of P (q) at the break−even point? Explain why this makes sense.
6)
Fuller and coworkers (1997) estimated the Costs of a medium-sized cotton gin plant as
shown in the following table:
2
4
6
8
10
12
15
20
x
217.6
271.12
325.5
380.8
437
494.16
581.7
732
y
x represents the annual number of bales produced in thousands.
y represents the total cost in thousands of dollars.
a.
Determine the best-fitting quadratic Trendline for this data and report the resulting Cost
Function.
The study noted that revenue was $63.25 per bale. At what level will production break
even? Please show all algebraic work for this problem and state your final result in a
complete sentence with appropriate units. (See previous problem for the definition of the
break−even point. You will need the quadratic formula to solve this problem.)
What is the Profit function?
What Profits do they make if they sell 7 thousand bales of cotton?
b.
c.
d.
Chapter 4 – Page 97
Answers to Selected Problems 1)
a.
b.
c.
d.
e.
f.
Suppose the demand function is D(q ) = −1.5q + 300 .
$150
$75
About 147.
R(q ) = −1.5q 2 + 300q
$15,000
$11250
2)
3)
a.
b.
c.
d.
e.
Let the Demand function be given by D(q) = −1.5q + 300 . Suppose fixed cost are
C 0 = 3600 and let the variable costs be $90 per unit.
C (q ) = 3600 + 90q
$10,350
$14,062
$3712.50
P(q ) = −1.5q 2 + 210q − 3600
4)
5)
a.
b.
c.
At about 80 units.
q = 79.654 units, but since q is rugs, we round to 80.
What is the value of P (q) at the break−even point? Explain why this makes sense.
6)
Chapter 4 – Page 98
Chapter 5: Limits Reading Questions 1.) What is a limit? Describe in your own words what limits are generally used for? Try to be
as specific and clear as you can.
2.) Do Problem #1 from the Homework section at the back of the chapter. Report what results
you got, including tables you used and the final result.
Section 5.1: Limits In this short chapter, we encounter a powerful tool which lies at the heart of calculus. While we
will not explore its full power and flexibility here, we will get enough exposure to it to help us
make conceptual progress in our class project. The introduction of limits into our toolbox also
officially pushes us into the study of "the calculus," as it’s known in the world of mathematics.
Now that we have a Profit function, the next step in our Project is to find a more precise value for
q that represents where our profits will be maximized. The graph gives us a rough and ready idea
of what that value is, but it would be nice to have more precise, sophisticated, and universal
method for finding where a function has a maximum (or a minimum).
We are often asked to find the value of a function f(x) when x = a, some particular value. This is
called evaluating a function. When we are instead interested in examining how the function is
behaving near x = a, as opposed to when x = a, we use the idea of "limit" instead.
Example 33 2 x 2 − 3x − 2
has no meaning when x = 2. (Why not?) What happens to
x−2
f (x) when x is near 2?
The function f ( x) =
Solution:
We start by evaluating f at several numbers that are "very close" to x = 2, as in the
following table:
x 1.99 1.999 2 2.0001 2.001
f(x) 4.98 4.998 − 5.0002 5.002
This table suggests that as x gets closer and closer to 2, from either direction, the
corresponding value of f (x) gets closer and closer to 5.
Actually, you can play with the numbers a bit and pretty much convince yourself that
the values of f (x) can be made as close as you want to 5 by taking values of x that are
sufficiently close to 2. In calculus language, we would say "the limit of f (x) as x
approaches 2 is the number 5." The formal notation that corresponds to this statement
is this:
Chapter 5 – Page 99
lim f ( x) = 5 or lim
x→2
x→2
2 x 2 − 3x − 2
= 5.
x−2
Visually, if we graph f (x) , we get the following, which looks a lot like a line:
As x gets closer to 2 from either side (left or right), the graph appears to want to
move closer and closer to a y value of 5. Note that even though the graph may
appear to indicate that the function has meaning when x = 2, it does not.
y
6
4
2
x
0
1
2
3
Can you come up with an
explanation for why this
function looks like a line?
Think of yourself as a dot on the graph. As you move along the graph from the left side
and get closer to x = 2, your height off the x axis, the y value, approaches 5. The same is
true as you move towards x = 2 from the right side, except in that case your height
would be decreasing towards y = 5. That's the idea of a limit! If moving along the graph
from either direction takes you to the same y value, then we say the function has a limit
at the x value in question and the limit value can be specified. ■
At this point, we are ready to “define” what we mean by a “limit.”
Chapter 5 – Page 100
Limit of a Function
Let f be a function and let a and L be real numbers. Assume that f (x) is defined (exists) for
all values of x near x = a.
Suppose the following:
● The value of x is very close but not equal to a on both sides of a
● The corresponding values of f (x) are very close (and possibly equal ) to L
● The values of f (x) can be made arbitrarily close to L for all values of x that are close
enough to a.
Then the number L is the limit of the function f (x) as x approaches a which is written as
lim f ( x) = L
x→a
This definition of the limit is not formal because it's not entirely clear what we mean when we say
"near," "very close" or "arbitrarily close," nor have we defined those terms. The example we just
saw and the conditions we laid out above provide us with intuitive notions about what limits are,
but they should not be mistaken for a rigorous treatment of this important mathematical tool.
Example 34 If f ( x) = x 2 + x + 1 , what is lim f ( x) ?
x→ 3
Solution:
Once again, we make a table showing values of f (x) as x gets very close to 3.
2.9
2.99
2.999
3 3.0001
3.01
3.1
x
f(x) 12.31 12.9301 12.99993
13.0007 13.0701 13.71
The table seems to suggest that as x approaches 3 from either the left or right, f (x )
gets very close to 13.
Answer: We conclude that lim f ( x) = 13 .
x →3
In this case, note that the function f (x) is defined when x = 3 because
f (3) = 3 2 + 3 + 1 = 13 . In this case, the limit of the function is equal to the actual value
of the function. Compare this to the previous example where the limit of the function
existed but the function did not exist at the value of a being considered. ■
Chapter 5 – Page 101
Section 5.2: Finding Limits with Excel Building these tables and computing these functional values by hand is a bit time consuming,
especially with values like x = 3.0001. Let's do the same thing in Excel, allowing it to do most of
the computational work for us.
Example 35 Find lim
x →1
x 3 + x 2 − 2x
using Excel.
x −1
Solution:
First, note that this function is not defined at x = 1, so we can't just substitute x = 1 into
the function to find the limit.
We'll start by labeling Columns A and B "x" and " f (x) ," respectively.
In Column A, type numbers that are very close to 1, lie 0.9, 0.99, 0.999, 1.001, 1.01,
1.1. Make sure the numbers are in order so you can see any patterns that emerge on
either side of x =1. It can look like this:
In Cell B2, we'll type the following:
=(A2^3+A2^2−2*A2)/(A2−1)
Chapter 5 – Page 102
When we copy this formula down to the bottom of the list, we get the following:
First, notice the value of Cell B6. Why is Excel giving the #DIV/0! error?
Next, note what value f (x) seems to get close to from both directions as x gets close to
1 (but not exactly equal to 1). It certainly looks like f (x) is getting close to 3.
Answer: lim
x →1
x3 + x 2 − 2x
= 3. ■
x −1
Chapter 5 – Page 103
Chapter 5 Practice Problems Use Excel to find the following limits. Make sure you choose at least four values on both sides of
the relevant x value and copy/paste your table into a text document to show your results. You
should show about 3 or 4 decimal places of accuracy in your cells.
1)
lim
ln x
x −1
2)
lim
ln x − ln 3
x −3
3)
lim
x 3 − 3x 2 − x + 3
x−3
4)
lim
5)
x →1
x →3
x →3
e2x + e x − 2
x →0
e x −1
1
1
−
lim x + 3 3
x →0
x
x −5
x − 25
6)
lim
7)
1⎞
⎛
lim⎜1 + ⎟
x→∞
x⎠
⎝
8)
Challenge: Give an algebraic explanation that demonstrates why lim
x → 25
x
x3 + 5x 2 + 4x − 4
= −4
x → −2
x+2
Chapter 5 – Page 104
Answers to Selected Problems 1)
1
2)
3)
8
4)
5)
−1/9 = =0.11111
6)
7)
e
8)
Chapter 5 – Page 105
Chapter 6: Rates of Change and Marginal Analysis Reading Questions 1.) What is the difference between average rate of change and instantaneous rate of change?
Explain in your own words.
2.) What roles do secant lines and tangent lines play in this chapter?
3.) Describe in your own words what you think marginal analysis is about.
4.) Describe in your own words the main idea behind the "Forward-Backward Method" for
finding instantaneous rate of change.
5.) Suppose f ( x) = x 2 − 3 x + 2 . What is the average rate of change between x = 4 and x = 6?
Show work and calculations.
6.) Suppose f ( x) = x 2 − 3 x + 2 . Use the “Forward-Backward Method” with h = 0.01 to
estimate the instantaneous rate of change at x = 5.
7.) Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 6.1: Average Rate of Change and the Secant Line One of the key applications of calculus is determining how one variable (R) changes with respect
to another (q). For example, how does total Revenue change as the quantity of items sold goes up.
You may be thinking to yourself: "Well, it goes up!" But, how fast, or at what rate, does it go up?
When does the Revenue grow "quickly" instead of "slowly?" And for that matter, does Revenue
always go up or can it go down as q increases? In this chapter, we explore these kinds of
questions, and we begin with the concept of rate of change. For example, let's suppose we have
the following data on the total U.S. consumption of petroleum in millions of barrels per day.
Years Since 1950 0
5 10 15
20
25
30
35
Amount
6.5 8.5 9.8 11.5 14.7 16.3 17.1 15.7
If we run a trendline (third−degree polynomial) on this data, Excel gives the following result:
US Petroleum Usage (millions of barrels) for
Years After 1950
y = -0.0007x3 + 0.029x2 + 0.0775x + 6.8061
18
16
14
12
10
8
6
4
2
0
0
10
20
Chapter 6 – Page 106
30
40
The data seems to show that petroleum usage grew between 1950 and around 1980 (x = 30),
when it appears to have decreased. We would say that the rate of change of petroleum usage
between 1950 and 1980 was positive (usage went up), and then the rate of change became
negative in about 1980 (as usage went down).
We can be more specific than this, however. We can ask, "What is the rate of change of
petroleum usage between 1960 and 1980?" From the graph, we can see that between 1960 (x=10)
and 1980 (x=30), usage went up. Hence, the rate of change will be positive. But exactly what is
the rate of change, numerically speaking? To answer this question, let's more carefully define
what is known as the average rate of change.
Definition: Average Rate of Change
The average rate of change of a function, f(x), between the two points x = a and x = b is equal
to the rate of change of the line that connects these two points on the graph of the function.
That is, the average rate of change is the slope of the line that connects the two points.
Average rate of change =
f (b ) − f ( a )
.
b−a
The line connecting these two points is called a secant line.
This definition may seem a bit odd, but perhaps a picture and an example will help. Note:
f (b) − f (a )
is equivalent to the formula for slope that you are probably used to…
b−a
y − y1
m= 2
x 2 − x1
Can you see how they are the same?
Chapter 6 – Page 107
Example 36 Given the function f ( x) = x 2 + 3 x + 4 , what is the average rate of change of the function between
x = −4 and x = 2?
Solution:
First, let's look at a graph of the function:
y
10
x
-5
0
5
You can easily verify that f (−4) = 8 and f ( 2) = 14 . These two points are displayed as
dots on the graph. According to the definition, the average rate of change of
f (x) between x = −4 and x = 2 is the slope of the line that connects these two points.
Here is a picture of that line.
y
10
x
-5
0
5
Because we have two points on the line, computing the slope is easy. It's simply:
f (b) − f (a)
14 − 8
6 1
=
= =
b−a
2 − (−4) 6 1
Answer: The desired rate of change is 1. This means that for every change of 1 in the x
direction, we see a positive change of 1 in the y direction. ■
Chapter 6 – Page 108
Example 37 Given the function f ( x) = − x 2 + 4 x + 15 , what is the average rate of change of the function
between x = 1 and x = 5?
Solution:
First, we compute f (1) and f (5) . These can be verified to be f (1) = 18 and
f (5) = 10 . We then connect these two points with a secant line and compute the slope
of the secant line.
y
20
10
x
0
5
Before we do any computations, however, notice that the secant line is now sloping
downwards. This means that between x = 1 and x = 5, the function is dropping in value
and so its rate of change (slope) will be negative.
f (b) − f (a ) 10 − 18 −8
=
=
= −2
b−a
5 −1
4
Answer: The average rate of change is −2. This means that between these two points,
for every change of (positive) 1 in the x direction, we see a change of −2 in the y
direction. ■
Example 38 Let's go back to the scenario on petroleum usage in the U.S. that opened the chapter. Here are the
data we were given.
Years Since 1950 0
5 10 15
20
25
30
35
Amount
6.5 8.5 9.8 11.5 14.7 16.3 17.1 15.7
What is the average rate of change between 1960 and 1980?
Solution:
We will use the raw data given to us in the table indicating that when x = 10, y = 9.8
and when x = 30, y = 17.1
Chapter 6 – Page 109
This gives us:
f (b) − f (a) 17.1 − 9.8 7.3
=
=
= 0.365
b−a
30 − 10
20
Answer: The units on the y variable are millions of barrels and the units on the x−axis
are years, so we would say that the average rate of change of petroleum usage between
1960 and 1980 is 0.365 million barrels per year. Note: It is very important that any
time we compute a rate of change we are careful to include the appropriate units, if they
exist. ■
Example 39 A company gathers data from test markets on a new product and its costs. The total costs to
produce this new item are estimated from the following table:
1000
2000
3000
4000
5000
q
8,771 $
12,765 $
20,132 $
42,059 $
65,883
C(q) $
Use the data to estimate the average rate of change between 2500 and 4500 units produced.
Assume an exponential model for the costs.
Solution:
Because we don't have direct data for q = 2500 or q = 4500, we need to estimate costs
at those numbers. We'll do this by running a Trendline, and get this:
y = 4768e0.0005x
$70,000
$60,000
$50,000
$40,000
$30,000
$20,000
$10,000
$0
0
1000
2000
3000
4000
5000
We can evaluate this function at 2500 and 4500. For example,
C (2500) = 4768e 0.0005(2500 )
≈ 16,642
Also, C (4500) = 45,238 .
Hence, we can compute the average rate of change to be:
Chapter 6 – Page 110
6000
C (b) − C (a) 45238 − 16642
=
b−a
4500 − 2500
28596
=
2000
≈ 14.30
Answer: The average rate of change is $14.30 per item produced. This means that
between 2500 and 4500 units, the average unit produced has a cost that is going up by
$14.30 per unit. ■
Section 6.2: Instantaneous Rate of Change and the Tangent Line There are times when the average rate of change is not necessarily what we are interested in
knowing. Instead, we may be curious about how a function is behaving at one specific point (as
opposed to what is happening between two points). In other words, we are trying to figure out
what is happening at one instant on the graph. For example, consider once again the graph of the
U.S. Petroleum usage, where the x-axis gives years since 1950 and the y-axis gives billions of
barrels of petroleum per day.
US Petroleum Usage (millions of barrels) for
Years After 1950
y = -0.0007x3 + 0.029x2 + 0.0775x + 6.8061
18
16
14
12
10
8
6
4
2
0
0
10
20
30
40
From the graph, it is pretty clear that in 1975 (x=25), petroleum usage is on the way up. Put
another way, its rate of change is positive. But what is the exact rate of increase on January 1,
1975? The real problem with this question is that as it stands now, we don't have enough
information to compute the slope of a line. We only have one point and any slope computation is
going to require two points. We can't just arbitrarily pick some other point on the graph, since
doing so would immediately imply an average rate of change, and we've indicated that's not what
we are trying to find.
In situations like this, the secant line connecting two points on a graph is no longer useful. There
is a line, however, that defines exactly what we're after. It's called the tangent line, which is the
line that touches the graph at the point in question but does not touch the graph anywhere else
Chapter 6 – Page 111
near that point. (I believe the word tangent comes from the idea to "kiss," so think of the line as
just barely kissing the curve at a single point.)
Here's a cleaned up picture of the petroleum graph, with the tangent line at x = 25 sketched in:
Millions Barrels
10
Years since 1950
0
20
40
This tangent line is unique for this point. That is, at x = 25, there is no other line that touches the
graph at x = 25 and nowhere else near that point. The original table where this data came from
indicates that at x = 25, y = 16.3. But that's only one point, not two. To compute the slope of this
tangent line, we need to know one more point on the line. We could do this by simply eyeing the
graph and estimating. For now, I hope you'll believe that this tangent line has a slope of 0.215,
which means that at exactly the time when 1975 arrived, daily petroleum usage was increasing by
0.215 million barrels of petroleum. Here’s our definition for instantaneous rate of change.
Definition: Instantaneous Rate of Change
The instantaneous rate of change of a function, f(x), at a single point x = a is the exact rate of
change of that function at that point and is found by determining the slope of the line tangent
to graph of the function at the point x = a.
Chapter 6 – Page 112
Example 40 The graph of f ( x) = x 2 + 3 x + 9 is shown. What is the instantaneous rate of change of f (x) at
the point x = 5?
Solution:
The only way to answer this question with the tools we have is to use a graph. Below,
you'll see the graph of the function and the tangent line.
y
100
80
60
40
20
x
-5
0
5
10
15
There are two points that are highlighted. The point (5,19) is on the parabola and on the
tangent line, while the point (8,40) is a second point on the line. (8.40) was chosen only
because it was pretty clear from the grid that this combination of x and y went together.
There is no easy way at this time to know for sure that (8,40) is actually on the line, but
it should be clear that it's close enough. Anyway, the slope of this line can now be
computed as:
y2 − y1 40 − 19 21
=
=
=7
x2 − x1
8−5
3
Answer: The instantaneous rate of change of f (x) at x = 5 is 7. ■
Realistically speaking, this is not a very efficient way to find instantaneous rates of change. We
will want a simpler, more efficient way to do it, and we'll explore that in a future chapter. Before
then, however, we want to point out a particular case of average rates of change that is closely
related to instantaneous rate of change.
Chapter 6 – Page 113
Section 6.3: Marginal Analysis and Instantaneous Rate of Change In business settings, there are a few particular rates of change that seem to be of most interest:
• How revenue changes according to how many items are sold. This is called marginal revenue.
• How costs change according to how many items are produced. This is called marginal cost.
• How profits change according to how many items are produced and sold. This is called
marginal profit.
In an economics context, we tend to think of these marginal functions as the instantaneous rates
of change of each respective quantity. They are sometimes referred to as “marginal change.”
Unfortunately, we've just seen how computing this kind of number is clunky and prone to
calculation error. In this section, we explore how to approximate the instantaneous rate of change
by using the average rate of change. We will develop two ways of doing this. The first is simply
computations of slopes of secant lines, while the second incorporates the idea of limits from the
previous chapter.
The central idea shared by both of these methods is that for most cases, the slope of a tangent line
at a point (which measures instantaneous rate of change) is very close to the slope of a secant line
between that point and a nearby point (which measures average rate of change).
The “Forward Method” for Computing Instantaneous Rate of Change In this first approach, we pick a point on the function and then move one unit to the right.7 Since
we move one unit forward, we call this the Forward Method. That new point becomes the second
point we use to draw a secant line. Hence, if we are interested in the point x = 2 , then we also
look at the point x = 3 . If we are interested in the point x = −5 , then we also look at the point
x = −4 (since –5 + 1 = -4). For this method, if we are interested in the point x = a, then we would
also look at the point x = a + 1.
Example 41 Consider the function f ( x) = − x 2 + 10 x − 19 . Estimate the instantaneous rate of change at x = 4.
Solution:
In this case, we have x = 4, so we need to also look at x = 4 + 1 = 5.
f (4) = 5 and f (5) = 6 . Thus, our instantaneous rate of change is going to be
approximated by:
f (5) − f (4) 6 − 5
=
= 1.
5−4
5−4
7
By convention, we always move to the right, and not to the left.
Chapter 6 – Page 114
A picture shows us what is happening:
y
8
Tangent line
Secant line
6
4
2
x
0
2
4
6
8
10
Certainly, the slopes of these two lines are not equal, but the slope of the secant line is
at least a rough estimate of the slope of the tangent line.
Answer: The instantaneous rate of change at x = 4 is about 1. ■
Example 42 Let's consider the familiar Petroleum example once again.
Let B ( x) = −0.0007 x 3 + 0.029 x 2 + 0.0775 x + 6.8061 be the function that gives the number of
millions of barrels of petroleum used per day x years after 1950. Use the Forward Method to
estimate the instantaneous rate of change at x = 25 (1975).
Solution:
Since x = 25, we need B (25) , which is computed to be 15.931. (We'll use the trendline
to get the function values rather than relying on the original table of data presented
earlier in the chapter.) We also need x = 26, one unit past 25. B(26) = 16.122 , which
you can verify if you so choose. Then the instantaneous rate of change at x = 25 is
approximately:
B(26) − B( 25) 16.122 − 15.931
=
26 − 25
1
= 0.191
Chapter 6 – Page 115
Here's a picture of this graph and the two relevant lines:
Millions Barrels
Tangent line
15
Secant line
10
5
Years since 1950
0
10
20
30
40
In this picture, it really does seem that the two lines have slopes that are very close in
value. In fact, if the two lines were not labeled, it might be hard to tell which one was
which.
Answer: The Forward Method tells us the instantaneous rate of change is about 0.1908
million barrels per day. (Don't forget to include the units!) Thus, usage is increasing at
this point by this rate. ■
Before we move to the second method, let’s summarize the first method:
Summary: Forward Method
Given a function f (x) and a point x = a, the instantaneous rate of change of f (x) at x = a is
roughly estimated by:
f (a + 1) − f (a)
= f ( a + 1) − f (a)
(a + 1) − a
Chapter 6 – Page 116
The “Forward‐Backwards” Method for Estimating Instantaneous Rate of Change The Forward Method (above) has us move forward one unit in the x direction from the
point x = a . The next plan of attack has us move both forward and backward from the point
where we are measuring the rate of change. For this reason, we will call it the “ForwardBackwards Method.” The amount of distance we move will vary, but to develop this method, let’s
start with an easy case where we move forward one unit, and we move backwards one unit.
Example 43 Let's look at the function f ( x) = − x 2 + 10 x − 19 , as we did earlier, and estimate the instantaneous
rate of change at x = 4 by going forward and backward by one unit.
Solution:
Since x = 4 is the point in question, we go back one unit to x = 3, and forward one unit
to x = 5. The functional values are f (3) = 2 and f (5) = 6 . Our estimate for
instantaneous rate of change is:
f (5) − f (3) 6 − 2 4
=
= =2
5−3
5−3 2
Notice that the Forward Method gave us a value of 1, which is significantly different.
Which one is better? Well, look at the graph of the new tangent line:
y
8
Tangent line
Plan 2 Secant line
6
4
2
x
0
2
4
6
8
10
The slope of the secant line under the Forward-Backwards Method looks to be very
close to the slope of the desired tangent line since the lines appear to be parallel.
(Remember, if two lines are parallel, their slopes are equal.)
Chapter 6 – Page 117
Compare this to the Result from the Forward Method:
y
8
Tangent line
Plan 1 Secant line
6
4
2
x
0
2
4
6
8
10
We conclude that in this example, the Forward-Backwards Method provides a better
estimate than the Forward Method did. ■
Example 44 Use the Forward-Backwards Method to estimate the instantaneous rate of change for our now
familiar petroleum function, B ( x) = −0.0007 x 3 + 0.029 x 2 + 0.0775 x + 6.8061 at x = 25 (1975).
Solution:
We need B (24) and B (26) . Calculations show that B (24) = 14.9733 and
B (26) = 15.4019 . Thus,
B( 26) − B (24) 15.4019 − 14.9733
=
= 0.2143
26 − 24
2
Chapter 6 – Page 118
The graph shows that these two points are relatively close to x = 24.
Millions Barrels
Plan 2 Secant line
Tangent line
15
10
5
Years since 1950
0
10
20
30
40
So we'll zoom in a bit to get a better idea of exactly what's happening around the region
of x = 4.
16
15.5
15
14.5
23.5
14
24
24.5
25
25.5
Chapter 6 – Page 119
26
26.5
There are indeed two lines in the picture. The top line is the tangent line at x = 25, and
the one on the bottom is the secant line between x = 24 and x = 26. Although it's hard to
see here, these lines are close to being parallel, indicating their slopes are close.
Hence, this method gives us an estimated instantaneous rate of change of 0.2143, which
looks like it's close to the actual value. ■
In the last example, we went forward and backward one unit, getting pretty good results. But
there's nothing magical about moving one unit. In fact, it might be obvious (once you think about
it) that if you go backwards and forwards by a very small distance (much smaller than 1) from the
point x = a , then the slope of the resulting secant line gives us a very good estimate of
instantaneous rate of change. We will let the small distance that we move away from our point be
called h.
Here's sort of a visual representation of what we're doing:
y
Tangent line
Secant line
x
2
1.999
h = 0.001
h = 0.001
2.001
The picture has obviously been blown up so we can see it in more detail. Note that the value of h
is "very very small." Also notice that the two lines are nearly parallel, indicating nearly equal
slopes. This in turn means that the slope of the secant line is a very good estimator of the
instantaneous rate of change (slope of the tangent line).
Now here's the best part! There's nothing that says that h has to be 0.001. It could be 0.00001, or
even 0.00000000000000000001. In fact, it could be arbitrarily small. We might make it small
enough so that our secant line slope is arbitrarily close to the tangent line slope. In other words,
we are exploring the behavior of the secant line's slope as h approaches, or gets very close to,
zero. This is the idea of limits from the previous chapter! If we could let h get "infinitely small,"
Chapter 6 – Page 120
we'd have a perfect value for the instantaneous rate of change. Since we can't actually do this,
we'll settle for sufficiently small values of h. Let's recap this method:
Summary: Forwards-Backwards Method
Given a function f ( x) and a point x = a, the instantaneous rate of change of f ( x) at x = a is
exactly equal to:
f (a + h) − f (a − h )
( a + h) − ( a − h)
f (a + h) − f (a − h )
= lim
h →0
2h
lim
h →0
The value of h is typically very small.
f (a + h) − f (a − h )
is sometimes called the difference quotient.
2h
If a value of h has been specified, then the instantaneous rate of change is essentially equal to
the difference quotient.
The expression
Let's return to our two previous functions and use Excel to refine our earlier results.
Example 45 Consider the function f ( x) = − x 2 + 10 x − 19 , as we did earlier. Estimate the instantaneous rate of
change at x = 4 with the Forward-Backward Method using at least three different small values of
h.
Solution:
Here's what a spreadsheet solution of this might look like:
Chapter 6 – Page 121
The formulas used to get these results are shown below:
Answer: As you can see, the difference quotient appears to have a value of 2.0, leading
us to the conclusion that the instantaneous rate of change of this function at x = 4 is 2.0.
It appears, at least in this case, that even a value of h = 0.1 is just as good as h = 0.001.
■
Example 46 Let's look one more time at the petroleum function.
Suppose B ( x) = −0.0007 x 3 + 0.029 x 2 + 0.0775 x + 6.8061 . Use the Forward-Backwards Method
to estimate the instantaneous rate of change at x = 25.
Solution:
Rather than using an Excel spreadsheet, let's just let h = 0.01. We're going to need
f ( 25 + 0.01) = f (25.01) and f (25 − 0.01) = f (24.99) . Very careful calculation gives
us f ( 25.01) = 15.21325 and f ( 24.99) = 15.20895 . Hence, our instantaneous rate of
change is essentially the difference quotient:
B(25.01) − B(24.99) 15.21325 − 15.20895
=
= 0.215
2(0.01)
0.02
Answer: The instantaneous rate of change is 0.215 million barrels per day. You can
check that this was the value we claimed to be the correct one at the beginning of the
chapter. ■
Chapter 6 – Page 122
Section 6.4: Focus on the Project The concept of average and instantaneous rate of change can help us explore how our project
functions are behaving. For example, we can ask how fast profits are increasing or decreasing
from the first “economies of scale point” (500,000 items) to the second “economies of scale”
point (1,100,000 items).8
For example, let’s look at our Revenue and Profit functions.
Example 47 Compute the average rate of change of Revenue between the first “economies of scale” point and
the second “economies of scale” point.
Solution:
We will use the Revenue function we found in a previous chapter:
R (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
We first need to find the total Revenue at 500,000 units ( q = 500 ) , and at 1,100,000
units ( q = 1,100 ) .
R (500) = 87.96805885 ⇒ 500,000 units bring in 87.968 million $ in revenue
and
R (1,100) = 81.79855657 ⇒ 1,100,000 units bring in 81.799 million $ in revenue
Therefore, the average rate of change is:
≈
81,798,556.57 − 87,968,058.85
= −10.28
1,100,000 − 500,000
Since the numerator is in dollars and the denominator is in items produced, this means
that on average, revenue is decreasing (the slope is negative) by $10.28 per item
produced. ■
8
See Chapter 2 for a review of what we mean by “economies of scale” points.
Chapter 6 – Page 123
Example 48 Use the Forward Method to find an estimate for the Instantaneous Rate of Change of Profits at
50,000 units past the first “economies of scale” point.
Solution:
The first “economies of scale” point for our class team data is 500,000 units. So, we are
interested in the point where we have 550,000 units.
Since we are using the Forward Method, we will need to compute the profit at
q = 550 (thousand) units, and q = 551 (thousand) units. Since we are past the first
“economies of scale” point, we need to use the second portions of the piecewise profit
function. From Chapter 4, we found that that portion of the profit function to be:
P2 (q ) = −0.00000011519q 3 + 0.00001501173q 2 + 0.09722734383q − 29.1
The profit values at the appropriate points (using Excel), are as follows:
P (550) ≈ 9.75162325 ⇒ Profits from selling 550,000 units are $9,751,623.25
and
P (551) ≈ 9.76065489 ⇒ Profits from selling 501,000 units are $9,760,654.89
We can now estimate the instantaneous rate of change at 550 thousand units:
9,751,623.25 − 9,760,654.89 9,031.64
=
= 9.03
551,000 − 550,000
1000
Hence, profits are increasing by approximately $9.03 per item when 550,000 thousand
units have been produced.
Chapter 6 – Page 124
If we want a more refined and accurate estimate of the instantaneous rate of change at 550,000
units, we can use the Forward-Backward method.
Example 49 Use the Forward-Backward Method to find an estimate for the Instantaneous Rate of Change of
Profits at 50,000 units past the first “economies of scale” point. Use h = 0.001
Solution:
Since we are using the Forward-Backward Method, we will need to compute the profit
at q = 549.999 (thousand) units, and q = 550.001 (thousand) units.
The profit values at the appropriate points (using Excel), are as follows:
P (549.999) ≈ 9.75161405 ⇒ Profits from selling 549,999 units are $9,751,614.05
and
P (551.001) ≈ 9.75163246 ⇒ Profits from selling 501,000 units are $9,751,632.46
We can now estimate the instantaneous rate of change at 550 thousand units:
9,751,614.05 − 9,751,632.46 18.41
=
= 9.21
549,999 − 550,001
2
Hence, using the Forward-Backward method, we estimate that profits are increasing by
approximately $9.21 per item when 550,000 thousand units have been produced. This is
about 18 cents different than our previous method. We can say that the marginal profits
at 550,000 units is about $9.21 per unit.
You are now ready do similar computations on your team’s data.
Chapter 6 – Page 125
Chapter 6 Practice Problems 1)
Find the average rate of change between (-3,2) and (4,-7).
2)
Find the average rate of change between (1,4) and (5,17).
3)
Find the average rate of change between (2.2,3.6) and (1.5,9.3).
4)
Find the average rate of change between (-3.5,6) and (4,-2.7).
5)
Let f ( x) = 3x 2 − 4 x + 1 . Find the rate of change between x = -1 and x = 3.
6)
Let f(x) = -2x3-x2+3x-1 Find the rate of change between x = 1 and x = 4.
7)
Let f ( x) = e 2 x . Find the rate of change between x = -2 and x = 0.
8)
Let f ( x) = 2 x − 3 . Find the rate of change between x = 6 and x = 14.
For problems (9) to (12),use the following scenario: In 1973, Braley and Nelson studied the effect
of price increases on school lunch participation in the city of Pittsburgh. The Revenue function
for the program was R (q) = −0.00381q 2 + 62.476q , where q is the number of lunches and R(q )
is in pennies/lunch.
9)
Find the average rate of change of Revenue between q = 5,000 and q = 10,000 units.
Include proper units and state your answer as complete sentence which indicates if Revenue
is increasing or decreasing.
10) Find the average rate of change of Revenue between q = 5,000 and q = 5,001 units. Include
proper units and state your answer as complete sentence which indicates if Revenue is
increasing or decreasing.
11) Find the average rate of change of Revenue between q = 10,000 and q = 15,000 units.
Include proper units and state your answer as complete sentence which indicates if Revenue
is increasing or decreasing.
12) Find the average rate of change of Revenue between q = 10,000 and q = 10,001 units.
Include proper units and state your answer as complete sentence which indicates if Revenue
is increasing or decreasing.
Chapter 6 – Page 126
For Problems (13) and (14), use the following graph to estimate the requested values. The graph
shows the growth of the US population, where x is years since 1790 and the population is
measured in millions of people.
US Population Growth
Pop (millions)
600
400
200
Years Since 1790
0
200
400
600
800
1000
13) Find the average rate of change of the population between 1890 and 1990. Include proper
units and state your answer as complete sentence which indicates if the population is
increasing or decreasing.
14) Find the anticipated average rate of change of the population between 1990 and 2190.
Include proper units and state your answer as complete sentence which indicates if the
population is increasing or decreasing.
For problems (15) to (16),use the following scenario: The Demand function for a product is given
by D(q ) = −0.0003q 2 − 0.04q + 23.56 where q is units sold and D(q ) is in dollars.
15) Find the average rate of change of Demand between q = 40 and q = 80 units. Include proper
units and state your answer as complete sentence which indicates what is happening to
Demand.
16) Find the average rate of change of Demand between q = 80 and q = 100 units. Include
proper units and state your answer as complete sentence which indicates what is happening
to Demand.
Chapter 6 – Page 127
For problems (17) to (18),use the graph of a Demand function to estimate the requested values.
On the graph, q is units sold and D(q ) is in dollars.
D(q)
20
10
q
0
50
100
150
200
17) Estimate the average rate of change of Demand between q = 50 and q = 100 units. Include
proper units and state your answer as complete sentence which indicates what is happening
to Demand.
18) Find the average rate of change of Demand between q = 80 and q = 100 units. Include
proper units and state your answer as complete sentence which indicates what is happening
to Demand.
19) A regional restaurant chain plans to introduce a new steak dinner. Managers have tested
various prices in their establishments and arrived at the following estimates for weekly sales
demand:
Price
$14.95 $19.95 $24.95 $29.95
# sold per week 2,800 2,300 1,600
300
a.
b.
Find the average rate of change in the quantity demanded between 1,600 units and 2,800
units. Include proper units and state your answer as complete sentence which indicates what
is happening to Demand.
Find the average rate of change in the quantity demanded between 300 units and 2,300
units. Include proper units and state your answer as complete sentence which indicates what
is happening to Demand.
Chapter 6 – Page 128
For Problems (20) to (23), use the given graph to estimate the instantaneous rate of change of
Revenue at the given point. On the graph, q is units sold and R(q) is thousands of dollars. Express
your answer in a complete sentence that includes proper units.
R(q)
120
100
80
60
40
20
q
0
200
400
600
800
1000
1200
1400
1600
180
20) Find the instantaneous rate of change at q = 200.
21) Find the instantaneous rate of change at q = 1200.
22) Find the instantaneous rate of change at q = 600.
23) Find the instantaneous rate of change at q = 800.
24) Given f ( x) = 2 x 2 − 3x + 5 , use the Forward Method to estimate the instantaneous rate of
change at the following points:
x = -2.
a.
b.
x = +2
c.
x = +3
25) Given f ( x) = x 3 − 10 x + 3 , use the Forward Method to estimate the instantaneous rate of
change at the following points:
x = -2.
a.
x = +2
b.
c.
x = +3
Chapter 6 – Page 129
26) Given f ( x) = 2 x 2 − 3x + 5 , use the Forward-Backwards Method to estimate the
instantaneous rate of change at the following points. Use h = 1 to do your computations.
x = -2.
a.
b.
x = +2
c.
x = +3
27) Given f ( x) = x 3 − 10 x + 3 , use the Forward-Backwards Method to estimate the
instantaneous rate of change at the following points. Use h = 1 to do your computations.
x = -2.
a.
x = +2
b.
c.
x = +3
28) Given f ( x) = 2 x 2 − 3x + 5 , use the Forward-Backwards Method to estimate the
instantaneous rate of change at the following points. Use h = 0.01 to do your computations.
x = -2.
a.
x = +2
b.
c.
x = +3
29) Given f ( x) = x 3 − 10 x + 3 , use the Forward-Backwards Method to estimate the
instantaneous rate of change at the following points:
x = -2.
a.
x = +2
b.
c.
x = +3
For problems (30) to (31),use the following scenario: In 1973, Braley and Nelson studied the
effect of price increases on school lunch participation in the city of Pittsburgh. The Revenue
function for the program was R (q) = −0.00381q 2 + 62.476q , where q is the number of lunches
and R(q ) is in pennies/lunch..
30) Estimate the instantaneous rate of change of the Revenue function at q = 5000 units as
follows:
a.
Using the Forward Method. Include units in your answer.
b.
Using the Forward-Backward Method with h=1. Include units in your answer.
c.
Using the Forward-Backward Method with h=.001. Include units in your answer.
31) Estimate the instantaneous rate of change of the Revenue function at q = 10,000 units as
follows:
a.
Using the Forward Method. Include units in your answer.
b.
Using the Forward-Backward Method with h=1. Include units in your answer.
c.
Using the Forward-Backward Method with h=.001. Include units in your answer.
Chapter 6 – Page 130
Answers to Selected Problems 1)
-9/7
2)
3)
-8.143
4)
5)
2
6)
7)
0.491
8)
0.25
9)
Revenue is increasing at the rate of $.053 per unit, or 5.3 cents per unit.
10)
11) Revenue is decreasing at the rate of $.328 per unit
12)
13) The population is increasing at the rate of about 1.9 million people per year.
14)
15) An increase of one unit sold decreases the price at which it sold by 7.6 cents.
16)
17) An increase of one unit sold decreases the price at which it sold by 8.5 cents.
18) An increase of one unit sold decreases the price at which it sold by 13 cents.
19)
a.
An increase of one unit sold decreases the price at which it sold by $0.00833 per dinner.
b.
An increase of one unit sold decreases the price at which it sold by $0.005 per dinner.
20)
21) Revenue is decreasing by about 0.26 thousand dollars per unit.
22)
23) Revenue is neither increasing nor decreasing. The instantaneous rate of change is zero.
Chapter 6 – Page 131
24)
25)
-3
a.
9
b.
c.
27
26)
27)
3
a.
3
b.
c.
18
28)
29)
2.0001
a.
2.0001
b.
c.
17.0001
30)
31)
a.
-$0.13278 per unit.
a.
-$0.13724 per unit.
a.
-$0.13724 per unit.
Chapter 6 – Page 132
Chapter 7: Numerical Derivatives Reading Questions 1.) Suppose you know that f ' (3) = 7 . What does that tell you about the function f(x)? Be
specific.
2.) Given the function f ( x) = 2 x 2 + x − 7 , find the value of the derivative when x = 4 and
interpret it. Let h = 0.01. Use the definition of the derivative in 7-1 to computer your result.
3.) Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 7.1: Derivatives To this point, when talking about average rate of change or instantaneous rate of change, we have
primarily talked about the marginal change or the marginal function. This approach serves us well
when we want concern ourselves with questions like "How are profits changing?" or "How is oil
consumption changing over time?" However, this is simply a special case of a more general
question: Given a function, f (x) , what mathematical tools can we use to precisely describe how
f (x) is changing at any given instant or point?
This is the fundamental question we will tackle in this chapter and it turns out to involve a
marginal function. In fact, the marginal function that we introduced in the previous chapter is
more generally referred to as the derivative of f (x) . That is, the derivative of a function is itself
a function that tells you how the function f (x) is changing…is it increasing, or decreasing, and
how fast? The formal definition for the derivative of a function uses limits and should look
familiar:
Definition: Derivative
Given a function, f (x) , the derivative of the function is defined to be:
f ' ( x) = lim
h →0
f ( x + h) − f ( x − h)
2h
The derivative of f (x) is the instantaneous rate of change of the function. The value of the
derivative at a particular point, x = a, corresponds to the slope of the line that is tangent to the
graph of f (x) at x = a.
In many cases, the value of this limit will be closely approximated by simply computing
f ( x + h) − f ( x − h)
for some very small value of h such as 0.01 or 0.001.
2h
Chapter 7 – Page 133
Because we cannot let h actually become 0 (why not?), we will compute the derivative in this
chapter by using a relatively small value for h, usually 0.01 or 0.001. Also, we should make sure
we're clear about how tangent lines play a role in all of this. The derivative is NOT a tangent line!
However, once we compute the derivative at a particular value of x, the value of the derivative
will be associated with a tangent line.
For example, at x = 2 , the function f ( x) = x 2 − 2 x + 3 has a derivative (instantaneous rate of
change) that is equal to 3. Here is a picture:
y
10
The tangent line is
not the derivative!
At x = 2, this function has a
tangent line with slope of 3.
That is, at x = 2, the derivative
of f (x) is 3.
x
5
-2
0
2
4
6
If we change the point we are looking at to x = −2, then we get a new result:
y
10
5
x
-2
0
2
4
6
In this case, the slope of the tangent line at x = −2 is −6. Hence, we would say that at x = −2, the
derivative of f (x) is −6.
Chapter 7 – Page 134
Saying all of this can get a bit tiresome, so we have notation that we use to help us out.
Notation: Derivatives
Given a function f (x) , it's derivative is denoted as f ' ( x) .
Now, instead of saying: "at x = −2, the derivative of f (x) is −6," we can instead just
say f ' (−2) = −6 .
Section 7.2: Computing Numerical Derivatives If a derivative is nothing more than marginal change or, more precisely, instantaneous rate of
change, then computing numerical values for derivatives is accomplished just like before. We
f ( x + h ) − f ( x − h)
.
choose a small value for h and then compute the difference quotient:
2h
Example 50 Given the function f ( x) = 2 x 2 + x − 7 , find the value of the derivative when x = 3 and interpret it.
Let h = 0.01.
Solution:
The "quick" way is to simply compute the following:
f ( x + h) − f ( x − h) f (3.01) − f (2.99)
=
2h
0.02
2
2
2(3.01) + 3.01 − 7 − 2(2.99 ) + 2.99 − 7
=
0.02
14.1302 − 13.8702
=
0.02
0.26
=
0.02
= 13
[
] [
]
Answer: The derivative at x = 3 has a value of approximately 13. ■
Technically, to compute the derivative in the example above, we'd need to take the limit of the
difference quotient as h approaches 0. Once again, however, the only method we have developed
to do this is numerical.
Chapter 7 – Page 135
Example 51 Use limits and Excel to compute the derivative of f ( x) = 2 x 2 + x − 7 at x = 3.
Solution:
For this approach to the problem, we want values of h to approach 0, so we explore
f ( x + h ) − f ( x − h)
what happens to
as h gets close to 0. A spreadsheet can be set up
2h
and might look like this:
Column F gives the value of the difference quotient as h gets very small. As you can
see, the values are all very close to 13, which we got previously. Since this is a lot more
work than the previous method, we are inclined to think that the first method is a more
efficient way to compute the derivative, at least numerically. ■
Can you think of a reason why Cell F8 is giving a Division by 0 error? What is its significance in
this context?
Chapter 7 – Page 136
Section 7.3: Focus on the Project This Focus will not require a lot of new calculations, but the material we cover is important. In
this chapter, we linked the definition of derivatives to the methods we used in the previous
chapters when we estimated the instantaneous rate of change. Essentially, the derivative of a
function is found by taking the limit of the Forward-Backward Method as h goes to zero. Hence,
we can get the Marginal Revenue and Marginal Profit functions with some careful (but tedious)
calculations. Fortunately, in the next chapter, we’ll discover a much more efficient way to deal
with derivatives.
Until then, let’s do one more calculation with our class data set.
Example 52 Find the Marginal Profits at the point that is 20,000 units past the second “economies of scale”
point for the class team data.
Solution:
The second “economies of scale” point for our class data is at 1,100,000 units (or
q = 1,100 ) units. So the point we are interested in is q = 1,120 , or 1,120,000 units.
Finding the derivative at this point typically involves choosing smaller and smaller
values of h, and computing a series of difference quotients. However, for our purposes,
if we take h = 0.001 , that will get us close enough until we have more accurate methods
in the next chapter.
This approach means we repeat a calculation similar to the one we saw at the end of the
previous
Since we are using the Forward-Backward Method, we will need to compute the profit
at q = 1,119.999 (thousand) units, and q = 1,120.001 (thousand) units.
The profit values at the appropriate points (using Excel), are as follows:
P (1,119.999) ≈ −63.00572865 ⇒ Profits from selling 1,199,999 units are negative
$63,006,313.90.
and
P (1,120.001) ≈ −63.00631390 ⇒ Profits from selling 1,120,001 units are
$63,005,728.65.
(Note: More decimal places of accuracy are shown in the numbers above to insure
what we can do the necessary computations below. All calculations were done in Excel,
however, to preserve accuracy.)
We can now estimate the instantaneous rate of change at 1,120 thousand units:
63,005,728.65-63,006,313.90
585.246
=−
= −292.62
1,120,001 − 1,119,999
2
Chapter 7 – Page 137
That number seems pretty extreme? What is going on here? A graph of what’s going on
here will be useful:
The slope of this
line is –292.62, so
profits are
decreasing by this
rate at this point.
Since 1,120 (thousand) units is far to the right of the Profit graph, the profits are
decreasing pretty dramatically at this point. This is obviously not a good place to be.
Using the Forward-Backward method, we estimate that profits are decreasing by
approximately $292.62 per item when 1,120,000 thousand units have been produced.
We can say that the marginal profits at 1,120,000 units is about -$0.29262 per unit.
You are now ready do similar computations on your team’s data.
Chapter 7 – Page 138
Chapter 7 Practice Problems For Problems (1) to (6), find the numerical value of the derivative of the function at the indicated
point. Use an Excel spreadsheet to do your computations. See Example 2 of Section 7-2 for an
example of how to set up the problems. For your answer, paste a few rows of your spreadsheet
into a file and print them.
1)
f(x) = 2x3+x2-1 at x = 0.5.
2)
f(x) = 2x3+x2-1 at x = 2.
3)
f ( x) = 2 x + 3 at x = 1.
4)
f ( x) = 2 x + 3 at x = 5.
5)
f ( x) = 0.1e.02 x at x = 100.
6)
f ( x) = 0.1e.02 x at x = 200.
7)
a.
b.
c.
8)
a.
b.
c.
Suppose R (q) = −0.4q 3 + 55q , where q is thousands of units and R(q ) is thousands of
dollars. Find the numerical value of the derivative of the function at the indicated point. Use
and Excel spreadsheet to do your computations. See Example 2 of Section 7-2 for an
example of how to set up the problems. Paste a few rows of your spreadsheet into a file and
print them and then interpret the meaning of your result, including appropriate units.
At 3,000 units.
At 5,000 units.
At 10,000 units.
Suppose P(q ) = −0.4q 3 + 25q − 100 , where q is thousands of units and P(q ) is thousands
of dollars. Find the numerical value of the derivative of the function at the indicated point.
Use and Excel spreadsheet to do your computations. See Example 2 of Section 7-2 for an
example of how to set up the problems. Paste a few rows of your spreadsheet into a file and
print them and then interpret the meaning of your result, including appropriate units.
At 3,000 units.
At 5,000 units.
At 10,000 units.
Chapter 7 – Page 139
Answers to Selected Problems 1)
2.5
2)
3)
0.44721
4)
5)
0.01478
6)
7)
a.
b.
c.
Revenue is increasing by $44.2 per unit.
Revenue is increasing by $25 per unit.
Revenue is decreasing by $65 per unit.
8)
Chapter 7 – Page 140
Chapter 8: Symbolic Derivatives and Graphs of Derivatives Reading Questions 1.)
2.)
3.)
4.)
5.)
6.)
7.)
Suppose f ( x) = 4 x 3 − 5 x 2 + 10 x + 3 . Show how to get f ' ( x) and give the final expression
for f ' ( x) .
What is a critical point of a function?
How are critical points related to optimization?
What is meant by "relative maximum?”
In your own words, describe the idea behind the First Derivative Test.
Use derivatives to find the instantaneous rate of change of f ( x) = 2 x 3 − 5 x 2 + 8 x + 11
when x = 3. Show work and calculations.
Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 8.1: Notation and Basic Rules of Derivatives In the last few sections we’ve been exploring the role of derivatives from numerical and graphical
perspectives. Most of the derivatives we’ve found so far have been found by looking at limits and
doing some pretty tedious, sensitive computations with small values like h = 0.01. In this chapter
our goal is to simplify this process considerably by finding rules for taking derivatives
symbolically.
Before we look at any rules, however, let’s review our notation for derivatives.
Notation: Derivative of a Function
Given a function, f(x), the derivative of f(x) is defined to be:
lim
h→0
f ( x + h) − f ( x − h)
2h
The derivative of f(x) will be denoted as f ́(x). We read this as “f prime of x.”
The derivative gives you the exact slope of the tangent line at a point on a graph.
Note the little ́ symbol which denotes the derivative. In some contexts, such as business, the
derivative of a function is often called the “marginal function.” For example, if we have a revenue
function, R(q ) , then we would denote the derivative of R as R ́(q), or MR(q ) . This is called the
“marginal revenue function.” The same idea holds for Marginal Costs, and Marginal Profits.
We are now ready to start exploring some rules for taking and finding derivatives of functions.
Chapter 8 – Page 141
We’ll start with the easiest rule.
The Constant Rule
If f(x) is a constant function, that is f(x) = c, the f ́(x) = 0.
This makes a lot of sense when we consider what a constant function looks like when it’s
graphed. For example, here’s the graph of f ( x) = 5 .
Clearly, this function is neither increasing nor decreasing. The rate of change at any point is flat,
or zero. Furthermore, imagine choosing any point on the graph and drawing a line tangent to it.
What do you get? Well, a horizontal line, of course, and the slope of any horizontal line is 0.
Since π is a constant, then if f (x) = π , then f ́(x) = 0.
The natural base, e, is also a constant, so if f ( x) = e , then f ́(x) = 0.
It doesn’t matter what the constant is, big or small, either. For example, if f(x) = 1,000,000,000,
then f ́(x) = 0.
Chapter 8 – Page 142
The next step up in functional complexity would be a basic linear function.
Suppose f ( x) = mx + b , or y = mx + b . In this case, the derivative is also pretty straightforward to
“compute.”
The Linear Rule
Given a linear function, f ( x) = mx + b , the derivative of f(x) is f ' ( x) = m .
Again, a specific example may make this pretty clear.
Example 53 Let f ( x) = 4 x + 2 . Find f ́(x).
Solution:
We’ll start by simply applying the rule, which says the derivative is equal to m, the
slope of the line. In this case, m =4, so f ́(x) = 4.
If we draw a picture, we can see why this makes sense.
10 y
x
-10
-5
0
5
10
-10
The Linear Rule states that the derivative of f(x) = m. In this case, the slope of f(x) is 4,
so that must be the derivative as well. We know that a line has a constant rate of change
equal to its slope, so no matter what point you choose on the line, the rate of change
(slope) is always the same. In this case, it’s 4. Likewise, think of picking a point on the
line and then drawing a tangent line at that point. The tangent line will be the same as
the original line, so the slope of the tangent line (derivative) is the slope of the original
line.
Answer: f '( x) = 4 . ■
Chapter 8 – Page 143
Example 54 Suppose you have a cost function C (q) = 32.3q + 2,000 . What is the derivative of C (q ) and what
does it mean?
Solution:
The Linear Rule says that the derivative is equal to the slope of the linear equation.
Hence, C ' (q) = 32.3 . This means that costs are increasing by $32.3 per unit no matter
what the value of q might be. $32.30 is the variable cost per unit in this example, while
$2,000 is the fixed cost portion of the cost function. ■
Example 55 The population of a small city is given by the function P (t ) = −300t + 20,000 , where t is time in
years since 1990. Find P ' (t ) and interpret its meaning.
Solution:
Because P (t ) is linear and the slope of P (t ) is -300, this means P' (t ) = −300 . Because
the slope is negative, it means the instantaneous rate of change is negative, so the
population has been decreasing by 300 people per year since 1990. ■
Section 8.2: Power Rule When functions get a little more complicated, so do the rules. For example, take f ( x) = x 2 . If we
look at a graph of this function we can plainly see that the slope of the tangent line, and therefore
the derivative, will change depending on at what point we choose to draw the tangent line.
y
10
m2
5m
1
x
0
5
-5
Above, we can see that the instantaneous rate of change at x = 1 (m1 ) is much smaller than the
instantaneous rate of change at x = 2 (m2 ) , since the slope of the tangent line at x = 1 is smaller
than the slope at x = 2.
Chapter 8 – Page 144
The next rule allows us to deal with powers of x (or any other variable.)
The Power Rule
Given a function of the form f ( x) = c ⋅ x n , the derivative of f(x) is :
f ' ( x) = c ⋅ n ⋅ x n −1
Notice that this rule says that to find the derivative of f ( x) = c ⋅ x n , we take the original power, n,
and multiply it by whatever constant, c, is in front of the function. Then, we reduce the original
power to n – 1.
Example 56 Find the derivative of f ( x) = x 2 .
Solution:
Notice that we can write this as f ( x) = 1 ⋅ x 2 , so that we recognize the constant, c, to be
equal to 1 while the power, n, is 2. Applying the rules gives us this:
f ' ( x) = 2 ⋅ 1 ⋅ x 2 −1
= 2 x1
= 2x
Answer: f ' ( x) = 2 x . ■
Example 57 Find the derivative of g ( x) = 5 x 4 .
Solution:
Applying the Power Rule, we take c = 5, and n = 4, and get the following:
g ' ( x) = 5 ⋅ 4 ⋅ x 4 −1
= 20 x 3
Answer: g ' ( x) = 20 x 3 . ■
This rule also applies to situations where the coefficients or powers are not nice and tidy.
Chapter 8 – Page 145
Example 58 Give a Revenue function of R(q ) = .243q 3 , find the derivative function, also known as the
Marginal Revenue function, R ' (q ) .
Solution:
In this case, c = 0.243 and n = 3 , so the Power Rules gives this:
R ' (q) = 3 ⋅ 0.243 ⋅ q 3−1
= 0.729q 2
Answer: MR( q) = R ' (q) = 0.729q 2 . ■
Example 59 Find the derivative of P (q) = 2.12q 3.44 .
Solution:
For this example, c = 2.12 and n = 3.44 .
P ' (q) = 2.12 ⋅ 3.44 ⋅ q 3.44 −1
= 7.928q 2.44
Answer: P ' (q) = 7.928q 2.44 . ■
One important pattern to notice in the last few examples is that the derivative is no longer a fixed
number. Instead, the derivative itself is an expression or, if you will, a function. This means that
we can plug numbers into these functions to get out particular values and we can even graph them
like we would any other function. This is helpful, because it gives us a quick and easy way to
determine the instantaneous rate of change of a function at just about any point we want. We will
explore this more later.
Example 60 Find the instantaneous rate of change of f ( x) = x 2 at x = 2.
Solution:
We saw from a previous example that f ' ( x) = 2 x , by the Power Rule. This general
derivative expression can be used to explore what is happening at x = 2.
f ' ( x) = 2 x
f ' ( 2) = 2 ⋅ 2
=4
Chapter 8 – Page 146
This tells us that when x = 2, the derivative is 4. So, we know that the function’s rate of
change at that point is 4. Another way of saying this is by stating that the slope of the
line tangent to f ( x) = x 2 at x = 2 is 4.
Answer: f ' (2) = 4 . ■
The previous example shows us exactly how the derivative can change as the value of x we
examine changes. If we let x = 10, then f ' (10) = 20 for the function f ( x) = x 2 . This tells us that
the instantaneous rate of change of f(x) is much larger at 10 than it is at 2. And this makes sense
geometrically because the parabola that corresponds to f(x) definitely gets steeper as we move to
the right on the graph.
Example 61 Suppose a revenue function is given by R(q) = .243q 3 . Find the instantaneous rate of change of
revenue when q is 8 and interpret what it means.
Solution:
We saw from a previous example that R' (q) = 0.729q 2 . We only need to substitute
q = 8 into this to get our result.
R' (8) = 0.729(8)
= 0.729(64 )
= 456.56
2
Answer: R' (8) = 456.56 , which means that when 8 units have been sold, the total
revenue is increasing by $456.56 per unit sold. ■
Section 8.3: Derivatives Term‐by‐Term The rules we’ve looked at so far have allowed us to take derivatives of simple, single-term
functions. What if our function looked like this?
f ( x) = 3x 5 − 0.25 x 4 + x 3 + 2 x − 1
First of all, notice that this function is simply the sum of several component terms of the form
cx n . If we can take each of these terms, differentiate one by one, and then combine the results,
we’d be in good shape. Fortunately, that’s exactly what we can do.
Chapter 8 – Page 147
Here’s the rule that allows us do deal with this kind of situation:
The Addition Rule
Given a function of the form: f ( x) = g ( x) + h( x) + ... + j ( x) , then the derivative of f(x) is:
f ' ( x) = g ' ( x) + h' ( x) + ... + j ' ( x)
In other words, you can take and find derivatives “term by term.”
Notice that this term-by-term process only works when the terms of the function are added (or
subtracted) to each other. This process will NOT work if the terms are multiplied or dividing each
other.
Example 62
Find the derivative of f ( x) = 2 x 4 − 5 x 3 + 3x 2 − 11x + 1 .
Solution:
This function has five terms, including the constant term on the end. Let’s take each
term and “differentiate” it separately:
f(x) =
2x 4
− 5x 3
+ 3x 2
− 11x
+1
f ́(x) =
8x 3
− 15 x 2
+ 6x
− 11
0
Notice the − 11x term has a derivative of –11. This is true because you can think of
− 11x as a linear function, so it’s derivative is simply its slope…in this case, it’s –11.
Also, the last term, +1, is a constant, so the derivative is 0.
Answer: f ' ( x) = 8 x 3 − 15 x 2 + 6 x − 11 . ■
Chapter 8 – Page 148
Example 63 Find the derivative of g ( x) = 3x 4 / 3 + 2 x 2 − 8 x − 3 .
Solution:
Using the Addition Rule and the Power Rule, this derivative is relatively tame:
4
g ' ( x) = 3 ⋅ x 4 / 3−1 + 2 ⋅ 2 x1 − 8 − 0
3
1/ 3
= 4x + 4x − 8
Answer: g ' ( x) = 4 x1 / 3 + 4 x − 8 . ■
Example 64 In a previous chapter, we encountered the function:
B ( x) = −0.0007 x 3 + 0.029 x 2 + 0.0775 x + 6.8061
which gives the number of millions of barrels of petroleum used per day x years after 1950 in the
U.S. Use this function to find B ' ( x) when x = 10 and then interpret the number you get.
Solution:
The derivative of B(x) is found with the Power and Addition Rules:
B ' ( x) = −0.0007 ⋅ 3x 2 + 2 ⋅ 0.029 x + 0.0775 + 0
= −.0021x 2 + .058 x + .0775
Knowing the general derivative function, we can now find B ' (10) .
B ' (10) = −0.0021(10) 2 + 0.058(10) + 0.0775
= −0.0021(100) + 0.58 + 0.0775
= −0.21 + 0.58 + 0.0775
= 0.4475
We need to remember that B(x) has units of millions of barrels of petroleum and x has
units of years. Therefore, the units on this instantaneous rate of change are millions of
barrels per year.
Answer: After x = 10 years (in 1960), the U.S. use of petroleum is increasing by 0.4475
million barrels per year. That is, it’s going up by 447,500 barrels per year. (We know
it’s increasing because the derivative is positive.) ■
Chapter 8 – Page 149
Example 65 Suppose you have a profit function given by P (q) = − q 2 + 5q − 3 , where q is thousands of units
and P(q ) is millions of dollars. Find P' (1) and P ' (3) and interpret each one.
Solution:
We’ll start by finding the general derivative:
P ' (q ) = −2q + 5 − 0
= −2q + 5
We can now evaluate the derivative at q = 1 and q = 3.
P ' (1) = −2 + 5 = 3
P ' (3) = −2(3) + 5 = −1
Therefore, after 1 thousand units have been sold, profits are increasing by 3 million
dollars per every thousand units. This is equivalent to saying that profits are increasing
by $3,000 per unit (since a million divided by a thousand is a thousand).
Also, after 3 thousand units have been sold, profits are decreasing by 1 million dollars
per every thousand units, or decreasing by $1,000 per unit.
To see what’s going on here, let’s take a look at a graph of the Profit function.
4 $ (Mill)
Profits are
decreasing.
2
Profits are
increasing.
q (Th)
0
2
4
-2
From the graph, we can see that at q = 1 thousand units, the graph is on the way up, or
increasing. Note that if we were to draw a tangent line at q = 1, its slope would be
positive (+3 to be precise). Somewhere around q = 2.5 units, the profits reach a peak
and begin to decrease. At q = 3 thousand units, the profit function is decreasing and the
instantaneous rate of change is –1 (thousand $ per unit). Again, a tangent line at this
point would have a negative slope. ■
Chapter 8 – Page 150
Section 8.4: Graphs of Derivatives All of the derivatives we’ve been looking at can themselves be thought of as stand-alone
functions. As such, they can be graphed. Doing so gives us interesting insight into how the
original function behaves. Let’s explore this in a little more detail by using a simple example.
Let a revenue function be given by R (q) = −4q 3 + 8q 2 + 3q , where q represents thousands of units
sold, and R(q ) is measured in thousands of dollars. What do we know about the
We begin by graphing the Revenue function on a limited domain, from x = 0 to x = 2 :
Notice that around q = 1.5 thousand units, the Revenue function appears to reach a peak. Before
that point (to the left), the slope of a tangent line will be positive. For example, at q = 1 thousand
units, the slope of the tangent line is positive, as we can see below.
Positive
slope
Chapter 8 – Page 151
An important observation for us to make here is that a positive slope corresponds to a point on the
graph where that function is increasing. That is pretty clear from the picture above. In other
words, when the derivative is positive, the function is increasing.
After that point (to the right), the slope of a tangent line will be negative. For example, at q = 1.8
thousands units, the slope of the tangent line is negative, as we can see below.
Negative
slope
Again, we can observe that when a negative slope corresponds to a point on the graph where the
function is decreasing. In other words, when the derivative is positive, the function is increasing.
Now, what happens to the slope of the tangent line at the peak? Can you guess?
Here is a picture of what that tangent line would look like:
Chapter 8 – Page 152
As you can see, this tangent line is horizontal, and we know from basic algebra that the slope of
any horizontal line is zero. At this point on the graph, is the function increasing, or is it
decreasing? Well, it’s doing neither! It’s leveled off at that instant/point. So, we make the general
observation that at a peak such as this, when the slope is zero, the function is neither increasing
nor decreasing. Let’s summarize these ideas all at once:
Relationship: Functions and Marginal Functions
When a function is increasing (going up), its derivative is positive (above the x axis).
When a function is decreasing (going down), its derivative is negative (below the x axis).
When a function is neither increasing nor decreasing, its derivative is potentially zero.9
The last fact in the box above gives us a hint about how we will locate peaks (and valleys) on
graphs. We’ll see that soon, but first let’s continue to explore more about how the graph of a
derivative is related to the original function.
Let’s continue to explore the revenue function given by R (q) = −4q 3 + 8q 2 + 3q .
Example 66 What is the marginal revenue function if R (q) = −4q 3 + 8q 2 + 3q and what does its graph look
like?
Solution:
The derivative of R(q ) is
R '(q) = −4 ( 3) q 2 + 2 ( 8 ) q + 3
= −12q 2 + 16q + 3
Since the derivative of the revenue function is also called the “marginal revenue
function,” we can say MR( q) = −12q 2 + 16q + 3 .
9
We say “potentially” because this is not always true. It is in many cases, but we need to be careful since
it’s not always true. Can you draw a function where this is not true?
Chapter 8 – Page 153
When we graph this quadratic function, we get the following:
Note that both the Revenue function and the Marginal Revenue (MR) function are
graph together.
BEFORE you read the next example, look at those two graphs and see if you can take
what you’ve read up until now and describe how they are related to each other!!!
Example 67 If R (q) = −4q 3 + 8q 2 + 3q , how is the graph of R(q ) related to the graph of MR (q ) ?
Solution:
To answer this question, let’s mark the peak of the Revenue function with a vertical
line.
Chapter 8 – Page 154
Notice that the peak of the Revenue function corresponds to where the Marginal
Revenue function crosses the horizontal axis, which is precisely where the Marginal
Revenue function is equal to zero! Hence, the Revenue function reaches a peak where
the derivative is zero, which is a general observation we’ve just made. Namely,
What about to the left of the peak? Notice that the graph of the derivative (MR) is
always above the horizontal axis to the left of the peak. Another way of saying this is
that the derivative is always positive in that region. (If a graph is above the horizontal
axis, its value is positive.) This corresponds precisely to where the original Revenue
function is increasing!
The revenue
function is
increasing to the left
of q = 1.5
The derivative (MR)
is positive to the left
of q=1.5
This is another one of our general observations. Namely, when a function is increasing
(going up), its derivative is positive (above the x axis).
Chapter 8 – Page 155
You can probably guess what happens to the right of the peak. The figure below shows
this:
The derivative (MR)
is negative to the
right of q=1.5
The revenue
function is
decreasing to the
right of q = 1.5
Notice that the graph of the derivative (MR) is always below the horizontal axis to the
right of the peak. Another way of saying this is that the derivative is always negative in
that region. (If a graph is below the horizontal axis, its value is negative.) This
corresponds precisely to where the original Revenue function is decreasing!
Example 68 The following graph shows the derivative, f '( x) , of a function, f ( x) . Describe how the function
f ( x) is behaving. That is, when is the function increasing? When is the function decreasing?
Where, if at all, does the function have a peak or valley?
Chapter 8 – Page 156
Solution:
We make use of our previous observations. Namely, a function is increasing if it’s
derivative is positive, and it’s decreasing if the derivative is negative.
On the derivative graph shown, we see that there are two regions where the derivative is
negative, and one where it’s positive. The first region, Region I shown below, goes
from −∞ to about -0.8. In this region, the derivative is negative and so we know that
f ( x) in increasing in that region.
Region I:
f '( x) is negative in
this region
Region III:
f '( x) is negative in
this region
Region II:
f '( x) is positive in
this region
In Region II, from x = −0.8 to x = 0.8 , the derivative is positive, and so that means the
function f ( x) is increasing in that region. Finally, in Region III, from x = 0.8 to ∞ ,
the derivative is once again negative, so the function goes back to decreasing.
This means that f ( x) is decreasing from x = −∞ to x = −0.8 , then f ( x) is increasing
from x = −0.8 to x = 0.8 , and then f ( x) is decreasing again from x = 0.8 to ∞ .
There are two points when the derivative is zero, so there are two points where we have
potentially either a peak (maximum) or valley (minimum). From what we know so far,
can you figure out what is happening at each of these two points. Try it. We’ll discuss
this more in the next section.
Chapter 8 – Page 157
Section 8.5: Optimization and Applications In the previous section(s), we saw how a function will have a maximum or minimum point when
the instantaneous rate of change (the derivative) is zero. In other words, if a function has a
maximum or minimum point, then it is probably true that the derivative is equal to 0. Now that
we have a way to find the general derivatives of functions, we can more accurately find where
functions actually reach maximum or minimum points.
We start with a definition:
Definition: Critical Point
A critical point of a function is a point where the (first) derivative is zero or undefined.
For our purposes here, we will say that a function's derivative is undefined at any point where you
cannot draw a unique tangent line at that point. Let's look at the picture below:
y
5
x
-5
0
5
There are four critical points on this graph. Two are points where the derivative does not exist.
These are the two to sharp points at x = −2 and x = +2. We can't draw a unique tangent line at
these points because the curve is not smooth at either point. Notice, however, that at x = −2, the
graph does have a minimum point and at x = +2, the graph has a peak. The other two critical
points are near x = −1 and near x = +1 where we see nice, smooth turning points for the graph.
Near x = −1, we have a maximum and near x = +1, we have a minimum. At both of these points it
is possible to draw a tangent line and each one will be a flat, horizontal line. Hence, each has a
slope of zero. This means that the derivative of the function at these two points is zero.
Chapter 8 – Page 158
More Definitions
The phrase relative maximum is used when we have a peak, a point that is higher than any
other nearby point on the graph. This is sometimes called a local maximum.
The phrase relative minimum is used when we have a low point, a point that is lower than
any other nearby point on the graph. Again, local minimum is sometimes used instead.
The phrase stationary inflection point describes a critical point that is neither a relative
maximum nor a relative minimum.
In the previous graph, we had two relative maxima and two relative minima.
For an example of a stationary inflection point, look at the following graph:
10 y
x
0
2
4
6
8
This graph has neither a local max nor a local min (no peak or valley). However, at x = 3, it has a
stationary inflection point. Here's why: if you were to draw a tangent line at x = 3, it would be
flat! Hence, at x = 3, the derivative of this function is 0 and so x = 3 is a critical point. By
definition, then, at x = 3, we have a stationary inflection point. This is, practically speaking, a
point where the graph momentarily levels off without reaching a high or low point.
We can now put several ideas and vocabulary together to get to the main point here. We want to
accurately locate local maxima and local minima points. To do so, we find all points where the
derivative of a function is zero and then check to make sure that we don't have a stationary
inflection point. Let's start with a simple example from the last section.
Chapter 8 – Page 159
Example 69 Find the local max or local min point for the function P (q) = − q 2 + 5q − 3 .
Solution:
We need to find when the derivative of P(q ) is zero. The derivative is
P' (q) = −2q + 5 , as we saw in the previous section. We now set this equal to zero and
solve for q:
− 2q + 5 = 0
− 2q = −5
− 2q − 5
=
−2
−2
5
q = = 2.5
2
This tells us that q = 2.5 is a critical point. We know that this is a parabola that opens
downwards, so we know we have a relative maximum point. ■
If we are not sure what the graph might look like, we need to do a little more work to know if we
have a maximum, minimum, or stationary inflection point. We can classify our critical points
accordingly by using what is called the first derivative test.
The First Derivative Test
Suppose f(x) is defined on the interval (a, b) and c is a critical point of f(x) in the interval (a, b) .
1.
2.
3.
If f ' ( x) > 0 for x near and to the left of c, and f ' ( x) < 0 for x near and to the right of c,
then f(x) has a relative maximum at x = c.
If f ' ( x) < 0 for x near and to the left of c, and f ' ( x) > 0 for x near and to the right of c,
then f(x) has a relative minimum at x = c.
If the sign of f ' ( x) is the same on both sides of c, then there is neither a relative max nor a
relative min at x = c, but instead we may have a stationary inflection point.
Note that the first condition basically says that if a function is increasing ( f ' ( x) > 0) and then
changes at x = c so that it's decreasing , then we have a relative maximum. This makes sense:
Relative max
Derivative is
positive…f(x) is
increasing.
Derivative is
positive…f(x) is
decreasing.
Chapter 8 – Page 160
Let's see how we might implement this test with an example.
Example 70 Find and classify all critical points of the function f ( x) = 2 x 3 − 3x 2 − 12 x + 4 .
Solution:
We start by finding the general derivative:
f ' ( x) = 3 ⋅ 2 x 2 − 2 ⋅ 3x 1 − 12 + 0
= 6 x 2 − 6 x − 12
Next, we find the critical points of this function by setting the derivative to 0 and
solving for x. Notice that we factor out a 6 first.
6 x 2 − 6 x − 12 = 0
6(x 2 − x − 2) = 0
6(x − 2 )(x + 1) = 0
⇒
x−2=0
or
x +1 = 0
x = 2, x = −1
This tells us we have two critical points and hence two possible points where there
could be a relative max, relative min or stationary inflection point. To figure out what is
going on, we put both of our critical points on a number line.
Let's focus on x = −1 for now. We need to check the value of the derivative to the left
and to the right of this point to see if it changes sign. To do this, we'll simply evaluate
the derivative at x = −2 (which is to the left of −1) and do the same at x = 0 (which is to
the right of −1).
f ' (−2) = 6(− 2 − 2)(− 2 + 1)
= 6(− 4 )(− 1)
= +24 > 0
So f ' ( x) > 0 to the left of c = −1.
f ' (0) = 6(0 − 2)(0 + 1)
= 6(− 2)(1)
= −24 < 0
So f ' ( x) < 0 to the right of c = −1.
Chapter 8 – Page 161
Checking the conditions of the First Derivative Test, we see that the derivative changes
from positive to negative, so the function increases to the left of x = −1 and then
decreases to the right of x = −1. This means there is a relative maximum at x = −1.
Now, we do the same thing for the other critical point, x = 2. To the left of x = 2, we
already know the derivative is negative since f ' (0) = −24 (see above). We'll use x = 3
to test the region to the right of x = 2.
f ' (3) = 6(3 − 2 )(3 + 1)
= 6(1)(4 )
= 24 > 0
So f ' ( x) > 0 to the right of c = 2.
Thus, to the left of x = 2, the function is decreasing and to the right of x = 2, it's
increasing. Hence, we have a relative minimum there.
We can actually show this on the number line graph. The + signs indicate a positive
derivative while the − sign's indicate a negative derivative.
Relative
Minimum
Relative
Maximum
++++++++++ − − − − − − − − − − − − − − − +++++++++
The function is
increasing.
The function is
decreasing.
The function is
increasing.
All of this makes visual sense when we look at the graph of this function:
10
y
x
-4
-2
0
2
4
6
-10
On the graph, you can see the relative max at x = -1 and the relative min at x = 2. ■
Chapter 8 – Page 162
Recap: Classifying Critical Points
1.
2.
3.
4.
5.
Find the derivative of the function.
Set the derivative equal to 0 and solve the resulting equation for the given variable. These
are your critical points.
Place all critical points on a number line.
Choose a number in each of the resulting regions on the number line and evaluate the
derivative at each point.
Use the criteria of the First Derivative Test to classify each critical point as a relative
maximum, relative minimum or stationary inflection point.
Caution: When testing points to the left and right of each region, do not make the mistake of
assuming the sign will automatically change. If a stationary inflection point exists, such a change
in the sign of the derivative will not occur.
Example 71 Let's consider the familiar Petroleum example once again. Let
B ( x) = −0.0007 x 3 + 0.029 x 2 + 0.0775 x + 6.8061
be the function that gives the number of millions of barrels of petroleum used per day x years
after 1950. The graph of this function look like this, you will recall.
US Petroleum Usage (millions of barrels) for
Years After 1950
y = -0.0007x3 + 0.029x2 + 0.0775x + 6.8061
18
16
14
12
10
8
6
4
2
0
0
10
20
30
40
Find the point in time when petroleum usage was the highest according to this model.
Solution:
First, we find B' ( x) :
B ' ( x) = −0.0007 ⋅ 3x 2 + 2 ⋅ 0.029 x + 0.0775 + 0
= −.0021x 2 + .058 x + .0775
Chapter 8 – Page 163
Set this equal to 0 and solve for x:
− .0021x 2 + .058 x + .0775 = 0
a = −.002
b = .058
c = .0775
Since there is no easy way to factor this (that I can see), we'll use the quadratic formula
with the values of , b, and c specified above.
− .058 ± .0582 − 4(− .002)(.0775)
2(− .002 )
− .058 ± .063119
=
− .004
= −1.278
or
= 28.896
x=
Since x = −1.278 makes little sense in this context (x being time), we discard that
numerical solution and keep 28.896.
Answer: Hence We can say that 28.896 years after 1950, usage reached a peak. This
would be 327 days into 1978, or roughly the end of November of 1978. ■
Example 72 A company manufactures and sells x transistor radios per week. The weekly cost function is
C ( x) = −5, 000 + 2 x and the weekly revenue function is R( x) = 10 x − 0.001x 2 (dollars). How
many radios do they need to sell to maximize their profits?
Solution:
We first need to find the profit function:
P ( x) = R ( x) − C ( x)
= 10 x − 0.001x 2 − (5000 + 2 x )
= 10 x − 0.001x 2 − 5,000 − 2 x
= 8 x − 0.001x 2 − 5,000
Next, we find the Marginal Profit function, MP( x) = P' ( x) :
P' ( x) = 8 − 0.002 x − 0
Setting this equal to 0 and solving gives the following:
Chapter 8 – Page 164
8 − 0.002 x = 0
8 = 0.002 x
x = 4,000
Thus, x = 4,000 is the only critical point. We'll test it just to make sure that it
corresponds to a relative maximum: To the left of x = 4,000, we'll find the value of:
P ' ( 0) = 8 − 0 = 8 > 0
To the right of x = 4,000 we'll find the value of:
P' (5000) = 8 − .002(5000)
= 8 − 10
= −2 < 0
The First Derivative Test tells us that x = 4,000 must be a relative maximum.
Answer: The company should sell 4,000 units (no more and no less) to maximize their
profits. (Can you think why of a reason selling more than 4,000 units would start to
decrease their profits?) ■
Example 73 Find and classify the critical points of the function: f ( x) = 2 x 4 − 4 x 3 + 2 x − 1 .
Solution:
First, we find the derivative:
f ' ( x) = 8 x 3 − 12 x 2 + 2
Setting this equal to 0 gives the following:
2(4 x 3 − 6 x 2 + 1) = 0
Unfortunately, we don't have an easy way to factor the inside of the parentheses and we
generally don't have methods to solve third−degree polynomials. Hence, there is no
algebraic solution to this problem.
Chapter 8 – Page 165
The best we can do is to estimate the answer from a graph:
y
5
x
-2
0
2
Answer: From the graph, it looks like we have a relative minimum near x = −0.4, a
relative maximum near x = 0.5, and another relative minimum near x = 1.4. That's all
we can really say at this point. ■
Chapter 8 – Page 166
Section 8.6: Focus on the Project At this point, we can now explore our project functions a little more carefully. From previous
Focus on the Project sections, we have the following list of project functions:
Project Functions
R (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
⎧21.6 + 0.115q if q ≤ 500
⎪
C ( q) = ⎨29.1 + 0.100q if 500 < q ≤ 1,100
⎪40.1 + 0.090q if q > 1,100
⎩
⎧= −.00000011519q 3 + 0.00001501173q 2 + 0.08222734383q − 21.6 if q ≤ 500
P(q ) = ⎨
3
2
⎩= −0.00000011519q + 0.00001501173q + 0.09722734383q − 29.1 ifq > 500
Example 74 At how many units is Revenue maximized?
Solution:
The marginal revenue function, MR( q) = R' (q) is as follows:
R' ( q ) = −0.00000034557 x 2 + 0.00003002346 x + 0.19722734383
If we set this equal to 0 and solve for q, we'll need the quadratic equation to get our
critical point(s):
0 = −0.00000034557 x 2 + 0.00003002346 x + 0.19722734383
q=
− .00003002346 ± .000030023462 − 4(− .00000034577 )(0.19722734383)
2(− .00000034557 )
Obviously, this is messy and care needs to be taken to find the solutions. Excel is a
good way to do this, and doing so gives q = 800.155 and q = −713.275 , the latter
being discarded.
Answer: Revenue is maximized at 800.155 thousand (or 800,155) units. ■
The cost function does not have a local max or a local min, since it's linear, leaving only the
Profit function, which is the main focus of this project. Recall that the graph of the Profit
function looks like this:
Chapter 8 – Page 167
$ (millions)
10
P2
5
P1
q (thousands)
0
200
400
600
800
From this graph we can see that we need to consider the profit function for q greater than 500.
Hence, in the following equation, we only need to worry about the second entry in the piecewise
function.
⎧= −.00000011519q 3 + 0.00001501173q 2 + 0.08222734383q − 21.6 if q ≤ 500
P(q ) = ⎨
3
2
⎩= −0.00000011519q + 0.00001501173q + 0.09722734383q − 29.1 ifq > 500
The marginal profit function of interest here, MP(q ) = P' (q ) , is equal to:
MP(q ) = −0.00000034557 q 2 + 0.00003002346q + 0.09722734383
We won't do any computations here, but setting this equal to 0 and then using the quadratic
formula to solve, we get q = 575.644 and q = −488.763. Again, only the former makes sense,
telling us that profits are maximized at 575,644 units sold and produced. Notice this gives us
pretty close to an exact answer and does not require that we estimate from the graph. Our
previous methods and work has not allowed us to do this.
This effectively answers the key question of this Project: How do we maximize profits? In the
next chapter, we'll explore a way to use Excel to do the grunt work in the algebra for us (ditching
the quadratic formula).
Chapter 8 – Page 168
Marginal Profit, Marginal Revenue, and Marginal Cost, Linked Together There is one more interesting thing to explore before we leave this important chapter. Previously,
we looked at the relationship between the graph of a function and the graph of its derivative.
Here, we explore how the marginal graphs for profit, revenue, and cost, are related to each other
and to the original functions.
First, note that if we want to maximize profits, we take the derivative of the Profit function, set it
equal to zero, and solve for q. But, this is the same as setting the Marginal Revenue function
equal to the Marginal Cost function. Why? Well, follow these steps to see what’s going on:
P = R−C
P ' = R '− C '
MP = MR − MC
MP = 0
8
MR − MC = 0
MR − MC + MC = 0 + MC
MR = MC
Profit = Revenue minus Costs
Take the derivative of both sides of this equation.
Write the derivatives in terms of marginal notation.
If the Marginal Profit function on the left side of the
equation is equal to zero, then the right side is also equal
to zero. This means MR − MC = 0
Add MC to both sides of the equation.
Simplify to get Marginal Revenue equal to Marginal
Cost.
This sequence of logic tell us that if Profits are maximized, then the Marginal Profit function is
equal to 0 AND Marginal Revenue is equal to Marginal Cost. This is a basic, well known “fact”
in the economics theory and at some point, you’ll see it again if you’re a business major.
This fact also translates nice graphically.
If we take derivatives of all of our class project functions, we get the Marginal Profit, Marginal
Revenue, and Marginal Cost functions.10
Now, if we graph those together, we get the following:
10
Graphmatica will compute and graph derivatives for you. Look for the Calculus menu. Make sure the
function you want to differentiate is selected. Then choose the “Find Derivative” command. Watch what
happens! ☺
Chapter 8 – Page 169
Marginal Change
0.2
MR = MC
when
q= 575.664
MR
MC
0.1
Units (K)
0
200
-0.1
-0.2
400
MP = 0
when
q = 575.664
600
800
MP
1000
1200
Why are
these flat?
MR = 0
when
q ≈ 800
-0.3
Since the Revenue function has just one piece, the Marginal Revenue graph has just one piece as
well. But the Cost and Profit functions are both piecewise with three pieces each, so the Marginal
Cost and the Marginal Profit functions each have three pieces, as you can see above.
Some things to notice and questions to ponder:
1.)
2.)
3.)
4.)
5.)
6.)
The Marginal Profit function crosses the horizontal axis (is equal to zero) at
q = 575.664 (thousand units). If we traveled up vertically until we hit the Profit curve, we
would do so at its peak.
The Marginal Revenue graph intersects the Marginal Cost graph intersect (are equal to each
other) at q = 575.664 (thousand units). This, of course, is related to what we said
above…Marginal Profit is equal to zero when Marginal Revenue equals Marginal Cost.
To the left of the vertical dotted line, notice that the Marginal Profit function is always
positive, which, as we’ve seen before, means the Profit function is increasing in that region.
If you look at the original Profit function, you can easily verify this. (See graph below)
To the right of the vertical dotted line, notice that the Marginal Profit function is always
negative, which means the Profit function is decreasing in that region.
The Marginal Revenue function is equal to zero at round q = 800 (thousand) units, which
means Revenue is maximized for that quantity.
The Marginal Cost function is composed of three flat (horizontal) lines. Why is this? Also,
why does the Marginal Cost function ever cross the horizontal axis?
Chapter 8 – Page 170
Here are the original functions for you to use as a reference:
$ (M)
150
C
100
R
50
P
0
200
400
600
Units (K)
800
1000
1200
140
You are now ready to do (lots of) similar calculations on your own team’s project data.
Chapter 8 – Page 171
Chapter 8 Practice Problems For Problems (1) to (4), find the derivative of the given function and simplify it completely.
1)
f ( x) = 3x 2 − 8 x + 2
2)
g ( x) = 5 x 4 − 2 x3 + 7 x 2 + 2 x + 10
3)
m(t ) = 2.4t 4 − 12.2t 2 + 15
4)
k (v) = 3v3 / 2 + 4v1 / 2 + 7
For Problems (5) to (8), find the derivative of the given function and then evaluate the given
value.
5)
f ( x) = −3x 4 − x3 + 2 x 2 + 1 at x = 3
6)
h( x) = 2.5 x3 − 3.4 x 2 − 6 x + 1 at x = 2
7)
j (t ) = 3t 5 / 2 − 6t 4 / 3 + 3t at t = 1
8)
z (r ) = .0025r 3 − 2.335r 2 − 5.24r − 12.1 at r = 50
9)
In an extensive study of cost functions for 40 firms in Great Britain, it was found that if x is
the output in millions of units and y is the total cost in thousands of pounds sterling (a
monetary measurement), then C ( x) = 13.311 + 2.3401x − 0.02009 x 2 . Find an expression for
the marginal cost function and then compute the value of the marginal cost function at the
following values of x, including proper units.
x=5
x = 10
x = 15
a.
b.
c.
10) If the revenue, R, in dollars for a product is given by R ( x) = −2 x 2 + 24 x , where x is the
number of items sold:
a.
Find the marginal revenue for any x
b.
Find R ' (3) and interpret it’s meaning.
c.
Find R ' (10) and interpret it’s meaning.
Chapter 8 – Page 172
11) Using data from the files of the National Commission of Food Retailing concerning the
operating costs of thousands of stores, researchers showed that the sales expense as a
percent of sales S was approximated by the equation, S ( x) = 0.4781x 2 − 5.4311x + 16.5795 ,
where x is in sales per square foot and x is less than 7.
a.
Find S ' ( x) for any x.
b.
Find S ' ( 4), S ' (5), S ' (6), S ' (7) and interpret what is happening.
12) Researchers have found the cost function for direct materials in a furniture factory was
approximated by C (q) = 0.667 q − 0.00467 q 2 + 0.000151q 3 for q between 0 and 50 units.
a.
Find the marginal cost function for any q.
b.
Compute the value of the marginal cost function for q = 5, 10 and 20 units and then
interpret what is happening.
c.
Graph the cost and marginal cost functions on the same set of axes.
13) In 1997, researchers at Texas A&M University estimated the operating costs of cotton gin
plants of various sizes. A quadratic model of cost in thousands of dollars for the largest
plant is given by C ( x) = 0.026689 x 2 + 21.7397 x + 377.3865 , where x is the annual quantity
of bales in thousands produced. Revenue was estimated at $63.25 per bale.
a.
Find the marginal cost function.
b.
Find the marginal revenue function.
c.
Find the marginal profit function
d.
Find the value of MP(30) and interpret its meaning.
In Problems (14) to (21), find all critical points and classify them as relative maxima, relative
minima, or stationary inflection points.
14)
f ( x) = x 2 − 4 x + 1
15)
f ( x) = x 4 + 2 x 2 + 1
16)
f ( x) = x3 − 3 x + 1
17)
f ( x) = 8 x3 − 24 x 2 + 18 x + 6
18)
f ( x) = x 4 − 8 x 2 + 3
19)
f ( x) = 3 x5 − 20 x3 + 5
20)
f ( x) = x 2 − 4bx + a
21)
f ( x) = x3 − 3bx + a
Assume b > 0
22) If the revenue function for a firm is given by R ( x) = 10 x − 0.01x 2 , find the value of x for
which revenue is a maximum. Use derivatives and show all work.
Chapter 8 – Page 173
23) If the cost function for a firm is given by C ( x) = 5 + 8 x and the demand equation is
D( x) = −0.2 x + 16 , find the value at which profits are maximum. Use derivatives and show
all work.
24) If the cost function for a firm is given by C ( x) = 9 x + 30 and the Revenue function
is R ( x) = x 3 + 36 x , find the value at which profits are maximum. Use derivatives and show
all work.
25) A researcher found that plant responses to phosphorus fertilizer was approximated by
f ( x) = −.057 − 0.417 x + 0.852 x1 / 2 , where f(x) is the yield and x the units of nitrogen. Find
the number of units of nitrogen that maximizes the yield. Use derivatives and show your
work.
Chapter 8 – Page 174
Answers to Selected Problems 1)
6x – 8
2)
3)
9.6t 3 − 24.4t
4)
5)
-339
6)
7)
5/2
8)
9)
a.
b.
c.
2,139 pounds sterling per thousand units.
1938 pounds sterling per thousand units.
1737 pounds sterling per thousand units.
10)
11)
a.
S ' ( x) = 0.9562 x − 5.4311
b.
S’(4) = -1.6063, S’(5) = -0.65, S’(6) = 0.3061, S’(7) = 1.2623. S(x) is decreasing initially
and then somewhere between x = 5 and 6, it attains a minimum and then begins to increase.
12)
13)
a.
b.
c.
d.
MC = 0.053389x + 21.7397
MR = 63.26
MP = -0.053389x + 41.5103
MP(30) = 39.91 which means profits are increasing by $39.91 per thousand bales sold when
30,000 bales have been sold.
14) Relative minimum at x = 2
15) Relative minimum at x = 0
16)
17) Relative max at x = 1/2, Relative min at x = 3/2
18)
Chapter 8 – Page 175
19) Relative max at x = -2, Relative min at x = 2, inflection point at x = 0
20)
21) Relative max at x = − b , relative min at x =
b)
22)
23) x = 20
24)
25)
x = 1.044 units of nitrogen.
Chapter 8 – Page 176
Chapter 9: Excel's Solver (Optional) Reading Questions 1.) What is the Solver Tool used for?
2.) What is the role of constraints when using the Solver Tool?
3.) Use the Solver Tool in Excel to find three different solutions to 4 x 3 − 8 x 2 + 3 x + 0.5 .
Section 9.1: Introduction In the previous chapter, we encountered some nasty equations to solve with the quadratic
formula. While they were solvable with care and caution, in this section we see how Excel can do
this for us using the Solver Tool. In addition, you'll remember the example from the previous
chapter when we could not solve the equation 2(4 x 3 − 6 x + 2 ) = 0 . The Solver Tool will also do
this for us as well.
Before your proceed any further, however, you may need to make sure that you have the Solver
Add−In installed.11 The following instructions are for Office 2003. To check if this is the case,
choose Add−Ins… from the Tools menu. Make sure the Solver box is checked, as pictured
below.
Make sure
this is
checked!
Section 9.2: Using the Solver Feature Let's start with a simple example, since that's the easiest way to learn how to use the Solver Tool.
11
Earlier versions of Microsoft Office may need these Add-Ins installed. Later versions may not require
this.
Chapter 9 – Page 177
Example 75 Solve f ( x) = x 2 − 8 x + 3 = 0 for x.
Solution:
Here's a graph of this function:
y
x
0
5
10
-10
-20
From the graph, it looks the graph has a y−value of 0 when x is around 0.5 and when x
is around 7.5.
In Excel, we need a cell for the value of x and one for f(x). Here's what a simple setup
might look like:
=B1^2-8*B1+3
We want this cell to
be 0.
Chapter 9 – Page 178
Our goal is to make Cell B2 = 0. Choose Tools → Solver… from the menu to get this
box:
Note that if you click in the "Set Target Cell:" box and then click once on Cell B2,
Excel will enter $B$2 for you. Also note that the "Value of:" button needs to be pressed
and a value of 0 is entered into the corresponding box (because we are solving Cell B2
for 0). Now notice that Excel needs to know how to go about doing this, and the way
you guide it is by telling it what cell to change so that Cell B2 can get to be 0. Of
course, it's Cell B1 that needs to change, so we enter that into the "By Changing Cells:"
box, as follows:
Now press the Solve button and Excel will actually change the value of Cell B1 to one
of the solutions and give you a chance to either keep the solution or restore all the cells
to their original values. In this case, Cell B2 has the value of 4.5E−07 which is Excel's
way of saying 4.5 × 10 −7 = .00000045 . While this is not exactly 0, it is "close enough"
by Excel's standards and for our purposes, we'll consider it adequate as well. It changes
Cell B1 to 7.605551, which is close to our rough approximation of 7.5.
Chapter 9 – Page 179
You've probably noticed that Excel is pretty dumb, and it only gives us one of the
solutions. We're smarter than that and know that two exist. It may turn out that the one
we want is not the one that Excel reports, so we need to know how to get the other one.
Here's one way: From the graph, we know that the other solution is less than 2. This
acts as a constraint on that solution, and we can tell Excel about this constraint.
Click on the Add button in the "Subject to Constraints" section of the Solver toolbox
and the following box will appear:
In this box, we want to tell Excel to make sure that the answer it gives back in Cell B1
is less than or equal to 2. So we fill out the box so it looks like this:
When we click on OK, we'll return to the original Solver toolbox, where we can click
on the Solve button to get the second solution.
The second solution is x = 0.394449, which is once again close to our estimate of 0.4. ■
You should be aware that you can add as many constraints as you need to. If you know a solution
is between 4 and 10, you can add two constraints, one telling it to find a solution greater than 4
and one telling you to find a solution less than 10. Constraints can also apply to the value you are
"solving" for. Consider the following example.
Chapter 9 – Page 180
Example 76 The size limitations on boxes shipped by your plant are as follows. (i) Their circumference is at
most 100 inches. (ii) The sum of their dimensions is at most 120 inches. You would like to know
the dimensions of such a box that has the largest possible volume. Let H, W, and L be the
height, width, and length of a box; respectively; measured in inches. We wish to maximize the
volume of the box, V = H⋅W⋅L, subject to the limitations that the circumference C = 2⋅H + 2⋅W ≤
100 and the sum S = H + W + L ≤ 120.
Solution:
Let's set up the problem in Excel. You may want to recreate this on your own to make
sure it all makes sense.
=2*A2+2*B2
=SUM(A2:C2)
=A2*B2*C2
I've entered somewhat reasonable values for H, W, and L to start. Our goal is to
maximize the volume in Cell F2 subject to the two constraints given to us, one related
to the Circumference and the other related to the Sum of the three dimensions. We now
targeting
move to the Solver Tool. We'll fill in a We
few are
of the
boxes the
to get us started:
We choose "Max" to
volume to be maximized.
maximize the volume.
Notice that we need to specify all
three of the cells (H,W,&L) so
Excel will change all 3 as needed to
maximize the volume.
Chapter 9 – Page 181
The next step is to enter our two constraints. We'll click on the Add button and enter the
first constraint as follows:
Here, we specify that cell D2, which has the Circumference calculation, needs to be less
than or equal to 100. We do the same for the constraint on the value of S in cell E2 and
get two constraints, as shown below:
When we click the OK button, Excel changes Cells A2, B2, and C2 to meet all of our
criteria and solve our problem for us:
Answer: To maximize volume subject to the constraints given, the height should be 25
inches, the weight 25 inches, and the length 70 inches. This will provide a maximum
volume of 43,750 cubic inches. No other dimensions will provide a larger volume. ■
Please note that in the previous example, the values we get for the dimensions of the box totally
depend on the constraints. If we change the constraints, then our results will also change.
Chapter 9 – Page 182
Example 77 In a previous chapter, we had to solve the equation 2(4 x 3 − 6 x 2 + 1) = 0 to find the critical points
of f ( x) = 2 x 4 − 4 x 3 + 2 x − 1 but found there was no algebraic way to do it. Solve this with Solver
Tool.
Solution:
We simply set up the spreadsheet like this:
=2*(4*B1^3-6*B1^2+1)
Using the Solver, we get x = 1.3666 with x = 1 in Cell B1. This is one of our three
critical points. Remember the graph.
y
5
x
-2
0
2
The solution x = 1.366 is the rightmost critical point. To find the other two, we can
either use constraints OR we could choose different guesses for Cell B1. If we start with
−0.5 in Cell B1, the Solver will give us x = −0.366 as the solution and if we start with x
= 0.4 in Cell B1, the Solver gives us x = 0.5 as answer. This example demonstrates a
different way to get all solutions: namely use the graph to make good guesses that the
Solver will use to find the closest solution. ■
Chapter 9 – Page 183
Section 9.3: Focus on the Project This section will be relatively short as there is little to do here but verify the results we have
already been getting from our graph and from our usage of derivatives.
Recall our Class Project function:
Project Functions
R (q ) = −0.00000011519q 3 + 0.00001501173q 2 + 0.19722734383q
⎧21.6 + 0.115q if q ≤ 500
⎪
C ( q) = ⎨29.1 + 0.100q if 500 < q ≤ 1,100
⎪40.1 + 0.090q if q > 1,100
⎩
⎧= −.00000011519q 3 + 0.00001501173q 2 + 0.08222734383q − 21.6 if q ≤ 500
P(q) = ⎨
3
2
⎩= −0.00000011519q + 0.00001501173q + 0.09722734383q − 29.1 ifq > 500
Let's use the Solver Tool to verify the value of q that maximizes profits. We already know this
should be q = 575.644 (thousand) units from the previous chapter.
Recall that our Profit graph looks like this:
$ (millions)
10
P2
5
P1
q (thousands)
0
200
400
600
800
From this graph, we see that it's the second piece of the Profit function that we need to be
concerned about, since the value of q of interest lies past q = 500. So, we want to maximize
P (q) = −0.00000011519q 3 + 0.00001501173q 2 + 0.09722734383q − 29.1
Chapter 9 – Page 184
With this Profit function, we know from a previous chapter that the Marginal Profit function is:
MP(q ) = −0.00000034557 q 2 + 0.00003002346q + 0.09722734383
There are two ways to do this with solver. The first way is just to use the Profit function and have
Solver find the maximum. We can set up the basic spreadsheet like this:
Cell B2 has the equation for Profits entered so that it changes as q changes. The same is true for
Cell B3 which has the Marginal Profit equation entered into it.
To maximize the Profit function, we would use the following entries into the Solver Tool:
Notice the "Max" button has been selected. Clicking on Solve changes the spreadsheet to this:
The value of q matches all of our previous results and Solver even gives us the maximum Profits
we can expect to earn given our Class Project Data.
The second way to do this is to have Solver find when MP, or Cell B3, equals 0, since Profits will
be maximized when MP = 0. Doing so gives the same result, however.
Chapter 9 – Page 185
Chapter 9 Practice Problems For Problems (1) to (9), graph each of the following given functions and then use Excel’s Solver
tool to answer the following questions.
1)
Find where the function f ( x) = 0.24x 2 + 3.2x - 3 has a minimum.
2)
Find where the function f ( x) = 2 x 3 + x 2 − 5 x + 17 has critical points.
3)
Find where the function f ( x) = 2 xe 2.5 x has a minimum.
4)
Identify and classify all critical points for f ( x) = 2 x 4 − x 3 − 5 x 2 + 3x + 1 .
e
2
⎛ x−μ ⎞
− 0.5⎜
⎟
⎝ σ ⎠
has a maximum if μ = 30 and σ = 11 .
5)
Find where the function f ( x) =
6)
What is the smallest possible value of z if z = x 2 + xy + 2 y 2 − 8 x + 3 y ?
7)
What is the smallest possible value of z, if z = x 4 + y 4 − 4 xy + 1 .
8)
Minimize z = 3x 2 + y 2 subject to the constraint x + y − 4 = 0 .
9)
Minimize z = 2 x 2 + xy + y 2 + x subject to the constraint 2 x + y = 3 .
σ 2π
10) A package mailed in this country must have length plus girth of no more than 108 inches.
Find the dimensions of a rectangular package of the greatest volume that can be mailed. Let
z be the length and let the girth be 2 x + 2 y .
For Problems (11) and (12) use the following information: A Cobb-Douglas production function
tells you how many units can be produced based on how much has been invested in labor costs
and capital costs.
11) Let P ( L, K ) = 15 L1 / 3 K 2 / 3 be a Cobb-Douglas production function, where P ( L, K ) is units
produced and L and K are labor and capital investments, respectively. Find the maximum
production level subject to the constraint 200 L + 100 K = 7,500,000 .
12) Let P ( L, K ) = 100 L0.2 K 0.8 be a Cobb-Douglas production function, where P ( L, K ) is units
produced and L and K are labor and capital investments, respectively. Find the maximum
production level subject to the constraint 200 L + 100 K = 7,500,000 .
13) A closed rectangular box whose volume is 324 cubic inches is to be made with a square
base. If the material for the bottom costs twice as much per square foot as the material for
the sides and top, find the dimensions of the box that minimize the cost of materials.
Clearly identify all of your variables and all equations you use.
Chapter 9 – Page 186
14) A firm has 3 separate plants producing the same product and has a contract to sell 1000
units of its product. Each plant has a different cost function. Let x, y, and z, be the number
of items produced at the three plants and assume the respective cost functions are:
C1 ( x) = 100 + 0.03 x 2
C 2 ( y ) = 100 + 2 y + 0.05 y 2
C 3 ( z ) = 100 + 12 z
Find the allocation of production in the three plants that will minimize the firm’s total cost.
Chapter 9 – Page 187
Answers to Selected Problems 1)
-6.66
2)
x = -1.095, 0.7613
3)
x = -0.4
4)
Relative mins and x = –1.09 and x = 1.17, and relative max at x = 0.294
5)
x = 30
6)
z = -23 when x = 5 and y = -2
7)
z = −1 at x = 1, y = 1, and z = −1 at x = −1, y = −1. (There are two minimum points.)
8)
x = 1, y = 3.
9)
x = 1, y = 1
10) z = 36, x = y = 18, with maximum volume of 11,664 cubic inches
11) x = 12,500, y = 50,000
12) x = 2000, y = 4000
13) 6 by 6 by 9
14) x = 200, y = 100, z = 700
Chapter 9 – Page 188
Chapter 10: Integration Reading Questions 1.) In your own words, describe what is meant by "area under a curve."
2.) In your own words, describe what is meant by "Consumer's Surplus."
3.) What is represented by the "Not Sold" region on the Demand curve?
4.) What does an Integral tell us?
5.) Suppose f ( x) = 2 x 5 − 3 x 4 + 3x + 7 . Find an antiderivative for f(x), showing how you got it.
6.)
7.)
8.)
Find the area under f ( x) = 2 x 2 − 3 between x = 2 and x = 6. Show your work
What does the Fundamental Theorem of Calculus tell you? How is it useful?
Read the Focus on the Project Section of the chapter and then briefly describe in your own
words what we will do in this chapter to help make progress on the team project.
Section 10.1: Area Under a Curve In the last few chapters, we have been focusing all of our attention on the idea of instantaneous
rate of change, how to measure it, how to see it on graphs, and how to use it to optimize
functions. This focus makes up one half of what is known in mathematics as "the calculus." The
other half of calculus is concerned with how to compute the "Area under a curve." To understand
what this means, consider the graph of the standard normal distribution, which you likely heard
about if you took Business Math Part I:
y
0.4
0.2
x
-3
-2
-1
0
1
2
3
By "area under the curve" we mean the area that is trapped between the graph and the x axis. You
may remember that this graph is a distribution of normally distributed data with a mean of 0 and a
standard deviation of 1. Also, the total area under this curve is equal to 1 since this is a
probability distribution curve. Finally, we learned that about 68% of all the data lies within one
standard deviation of the mean. Therefore, the 'area under the curve" that is shaded above is 0.68
units. In Business Math Part I, we did not talk about how this area was computed and when we
needed to do so ourselves, we used the NORMDIST or the NORMSDIST commands in Excel. In
this chapter, we'll essentially learn how these commands do these sorts of computations. But first,
let's take a look at a couple more examples.
Chapter 10 – Page 189
Here's a picture of the area under f ( x) = x 2 from x = −2 to −2. The total area represented by the
shaded region is about 5.33 units.
6 y
4
2
x
-3
-2
-1
0
1
2
3
As one last example, here is the "area under the curve" of f ( x) = x 3 from x = −3 to x = +3:
y
20
Positive
Area
x
-4
Negative
Area
-2
0
2
4
-20
The area here is symmetric so it might make sense that the total area is actually two times one of
the two wedges. However, any area that is below the x axis is counted as negative area. So in this
case, we actually say that the total area is 0! While that may seem a bit strange, it's merely a
convention that mathematicians have chosen. It is still possible to compute the "true" area of the
two shaded regions, but in this course, we will not be seeing any applications where we need to
take into account "negative area."
Chapter 10 – Page 190
We now define what we mean by area under a curve:
Area Under A Curve
We define the area under a curve to be the area trapped between the curve and the x axis. This
area is considered positive if it is above the x axis and it is assigned a negative sign if it is
below the x axis.
Section 10.2: Consumer Surplus There is a very convenient way to picture the revenue that results from selling q items with a
Demand function D for a good. Since R(q) is the product of q and D(q), it is the area of a
rectangle that is q units wide and D(q) dollars high. It is easy to picture such a rectangle under
the graph of the Demand function.
Demand Function
D(q)
D(q)
Revenue
q
q
Since q is a number of items and D(q) is the price in dollars per item, the area of any region under
the graph of the Demand function represents income from sales. The unit of such an area is
(items)⋅(dollars/items) = dollars. In particular, the area of the entire region under the graph of
the demand function and over the horizontal axis represents the total possible revenue for the
good.
The total possible revenue is the money that the producer would receive if everyone who wanted
the product bought it at the maximum price that he or she was willing to pay. This is the greatest
possible revenue that a seller or producer could obtain when operating with a given Demand
function.
Chapter 10 – Page 191
Demand Function
D(q)
Total possible revenue
q
We have seen that our revenue from selling a good at a single price that must be paid by all
buyers is represented by the area of a rectangle that is part of the region under the graph of the
Demand function. The remaining space under the graph of the demand function is partitioned
into two other regions. These represent income that is lost from selling at a single price.
Some buyers would have been willing to pay a higher price for the good than we charged. The
total extra amount of money that people who bought the good would have paid is called the
consumer surplus. It is represented by the area of the region under the graph of the Demand
function and above the revenue rectangle.
Demand Function
D(q)
Consumer's Surplus
Revenue
Not Sold
q
Chapter 10 – Page 192
Fixing a single price means that some people can buy the good for less money than they would
have been willing to pay otherwise. It also means that some potential customers do not buy the
good, because they feel that the price is too high. The total amount of this lost income, which we
will call Not Sold, is represented by the area of the region under the graph of the Demand
function to the right of the revenue rectangle.
Demand Function
D(q)
Not Sold
Revenue
q
We learned how to find the single price that maximizes the revenue that we receive from selling a
good. This is an important tool. However, the skilled marketer is always looking for creative
ways to claim some of the lost income that is represented by the other regions under the graph of
the Demand function. The first step in this direction is computing the dollar values of the Total
Possible Revenue, the Consumer Surplus, and the Not Sold region.
Example 78 Suppose a food chain is looking at pricing buffalo steak dinners. The restaurants’ Demand
function is given by D(q) = −0.0000018⋅q2 − 0.0002953⋅q + 30.19.
a.
What is D( 2300) ?
b.
About how much of the total possible revenue is realized when the dinner's are priced
accordingly? Show this on the Demand graph.
c.
Can the maximum possible revenue be determined from the demand graph? If so, how?
Chapter 10 – Page 193
Solution:
First, let's get a basic picture of the of the demand and revenue curves:
D(q)
30
20
10
q
0
1000
2000
3000
4000
D(q) and R(q)
x10^4
x10^4
q
0
1000
2000
3000
4000
Solution a:
D( 2300) is the price that they can expect to charge if they sell q = 2300 dinners.
D( 2300) = −0.0000018(2300 ) − 0.0002953(2300 ) + 30.19
= 19.99
2
Answer a: At $19.99, they can expect to sell 2,300 dinners.
Solution b:
The total revenue when 2,300 dinners is sold is easy to compute:
2300 dinners ×
$19.99
= $45,977
dinner
Chapter 10 – Page 194
This can be represented on the graph as rectangle:
D(q) and R(q)
30
20
10
Revenue = $45,977
q
0
2000
4000
Answer b: From the graph, it looks like about half of all the possible total revenue is
being realized at the price of $19.99.
Solution c:
If we wanted to find the value of q that corresponds to maximum revenue, we would
have to experiment with q until the area of the rectangle is the largest it can be. This is
obviously not easy to do. It is certainly easy to use the Revenue graph, which tells us
that maximum revenue actually appears to coincide with q = 2,300 units. If we were to
set the marginal revenue (derivative) function equal to 0 and solve for q, we'd get 2,310
units.
D(q) and R(q)
30
20
10
Revenue = $45,976
0
2000
q
4000
Answer c: Maximum revenue takes place at q = 2,310 units, yielding $45,976 in
revenue, but that is hard to read from the demand curve. ■
Chapter 10 – Page 195
Section 10.3: Estimating Area Under a Curve The remaining question is how to compute what these areas are in a relatively easy manner. The
way we do this is by breaking up the curve into small pieces and estimating the area of each piece
with a simple rectangle (areas of rectangles are easy to find), and then adding up all the
rectangles.
Let's start with an illustration of the concept and then we'll move on to a simple example to
illustrate the computational process.
Recall the normal distribution curve that looks like this:
y
0.4
0.2
x
-3
-2
-1
0
1
2
3
Let's say we want to find the area under the curve between −1 and +1, which we know to be about
0.68.
We'll start by taking the interval [−1,1] and breaking it up into 4 equal subintervals, each 0.5 units
long. We then take the midpoint of each interval and draw a rectangle with a height that
corresponds to the height of the curve at the midpoint. Here's a picture of that:
Although we won't show the calculations here, these four rectangles have a total area of about
0.69. You can see that each rectangle does a decent job of estimating the area under the curve for
Chapter 10 – Page 196
the subinterval to which it corresponds. We can make our estimate more accurate by increasing
the number of rectangles. Let's double to 8 subintervals and rectangles:
Hopefully, it's clear that the gaps between the rectangles and the curve are getting smaller, so
there should be less error with 8 rectangles. In fact, what we do is let the number of rectangles
head off toward infinity and then look at the limit of the area. With a LOT of rectangles, you get a
great estimate, as the graph with 64 rectangles shown below illustrates:
This almost looks like we've shaded the region when in fact the picture really has 64 very skinny
rectangles graphed with a resulting approximate area of 0.68. (Any more than 64 and we're not
really getting much improvement in our estimate, unless you want to go out several decimal
places.)
Let's now look at how the computations are done in an example like this.
Chapter 10 – Page 197
Example 79 Find the area under the curve f ( x) = 2 x −
x2
between x = 1 and x = 4, using n = 6 subintervals.
2
Solution:
Let's start with a picture of what we want to compute:
Since we are finding the area between 1 and 4, this gives us a total interval length of 4 −
1 = 3 units. Each of the 6 subintervals are equal in length, so each one is 3/6 = 0.5 units
long. This process should look familiar. Let [1,4] = [a,b], where a and b are the
beginning and end points of our interval, respectively. Let Δx be the length of each
subinterval which we have seen computed back in Chapter 1 as:
Δx =
b−a
n
With the value of Δx computed, we can specify the start and end points of the
subintervals. These are given below:
x0 x1 x2 x3 x4 x5 x6
1.0 1.5 2.0 2.5 3.0 3.5 4.0
Notice we have seven numbers to make up our 6 subintervals, the first of which starts at
1.0 and ends at 1.5. The second starts at 1.5 and ends at 2.0. We'll label the first value
x0, so that a = x0.
Each of these subintervals has a midpoint which we can compute by finding the average
of the subinterval's start and end point. For the subinterval starting at x0 and ending at
x1, we'll call it's midpoint m1. The other midpoints will be m2, m3, m4, m5, and m6. Notice
that we will always have the same number of midpoints as we have subintervals.
Chapter 10 – Page 198
Here is a list of the midpoints for this example:
m1
m2
m3
m4
m5
m6
1.25 1.75 2.25 2.75 3.25 3.75
x0 + x1 1 + 1.5
=
2
2
2.5
=
= 1.25
2
m1 =
Each midpoint is Δx =
0.5 units away from the
next and previous one!
x5 + x6 3.5 + 4
=
2
2
7.5
=
= 3.75
2
m6 =
The next step is to take each of these midpoints and figure out the value of f (x) at
each one. Each of these values is going to be the height of the rectangle for that
subinterval. For example, for the first subinterval, we find f (1.25) :
f ( m1 ) =
= f (1.25) = 2(1.25) −
(1.25)2
2
1.5625
2
= 1.71875
= 2 .5 −
The first rectangle is 1.71875 units high and 0.5 units wide. (The width of each
rectangle will always be Δx units.). This gives an area of (1.71875)(0.5) = 0.895375
square units.
We can do this for all our rectangles, numbering them from i = 1 to 6:
n
0
1
2
3
4
5
6
xn
1
1.5
2
2.5
3
3.5
4
Midpoint
f(xn)
Area
1.25
1.75
2.25
2.75
3.25
3.75
1.71875
1.96875
1.96875
1.71875
1.21875
0.46875
0.859375
0.984375
0.984375
0.859375
0.609375
0.234375
Sum
4.53125
Adding up these 6 areas give us a total approximate area of 4.53125 square units. If we
wanted to use summation notation, we could express this as follows:
6
Sum of rectangular areas =
∑ f (m )Δx
i
i =1
Chapter 10 – Page 199
Here's a picture of what we've done. You can see that there are some gaps that represent
the area in our estimate but it's a start.
2
1
1
1.5 2 2.5 3 3.5 4
Answer: The approximate area is 4.523125 square units. ■
You can see how these computations can be done in Excel by looking at the file called Area
Example.xls. The sheet called "Any n" shows the computations for n = 6, but you can change n to
other values to see how it affects the area estimate. If we let n be very large, say n = 1000, we get
an area of 4.5 units.
The process we just went through can be generalized for any value of n, any function f(x), and
any interval [a,b].
Approximating the Area Under a Curve with Midpoint Rectangles
Let f (x) be a "smooth and continuous" function. To estimate the "area under the curve" in
the interval [a,b] for n subintervals, compute the sum:
n
∑ f (m )Δx
i
i =1
In the equation above:
b−a
n
and
x + xi
m i = i −1
2
(the average of the endpoints of the subinterval).
Δx =
Chapter 10 – Page 200
Example 80 Find the area under the curve of f ( x) = 1 − e − x / 2 on the interval [2,6] using n = 20 subintervals.
Solution:
First, here's a picture of what we are seeking:
We will do this example in a worksheet called Cumulative.xls. We'll start with entering
our basic parameters and computing Δx.
=(B2-B1)/B3
Next, we find the value of our subinterval start and end points, as well as the
corresponding midpoints. Here's what we get:
a = x0
b = xn
Chapter 10 – Page 201
And here are the formulas used to get these results:
a = x0
b = xn
Our next step is to find all the f (mi ) values and the corresponding rectangle areas. In
Excel it looks like this:
a = x0
b = xn
Chapter 10 – Page 202
Again, here are the formulas used to get these results:
a = x0
Answer: The area between 2 and 6 is approximately 3.64 square units. ■
Section 10.4: Integrals The examples we have been working through only approximate the area under the curve. In many
ways, what we have been doing is analogous to what we did with secant lines when we were
estimating the instantaneous rate of change. We let the value of h get very small and the slope of
the secant line became increasingly close to the slope of the tangent line. In this chapter, we let
the number of rectangles, n, get very large and then the area of all the rectangles becomes
increasingly close to the actual area under the curve.
As before, we can invoke the notion of limits to define the area under a curve. In particular:
Area Under the Curve: The Integral
n
Area Under the Curve = lim
n→∞
∑ f (m )Δx
i
i =i
To specify our interval [a,b] and to collapse this notation a bit, we'll denote the area under the
curve as follows:
b
∫
Area Under the Curve = f ( x)dx
a
b
We call the symbol
∫ f ( x)dx the integral of f(x).
a
Chapter 10 – Page 203
b
Note above that we are saying that
∫
n
f ( x)dx = lim
n→∞
a
∑ f (m )Δx .
i
i =i
b
For
∫ f ( x)dx , we say this is "the integral of f(x) from x = a to b."
The dx is essentially playing
a
the role of Δx in the summation notation. (It's a bit more complicated than that, but we'll live with
that for now.) Notice that the integral symbol looks like a tall, thin S and that is by design. The S
stands for sum, as in summation.
It is of course highly problematic to try to let n be infinite in number…you can't literally add up
an infinite number of areas (just as you can't let h actually become 0 when estimating
instantaneous rate of change). However, the limit concept rescues us once again allows us get an
exact value of the area under the curve.
Example 81 6
⎛
x2 ⎞
Find the value of ⎜⎜ 2 x − ⎟⎟dx and interpret its meaning.
2⎠
2⎝
∫
Solution:
x2
between x =
2
2 and x = 6. We have no way (at this time!) to compute this exactly, the best we can do
is to estimate it with a large number of rectangles and take an educated guess at the
exact number. In the previous section, we say that with n = 1000, the area was 4.5, so
that's all we can say at this time. (It turns out this is the correct result, however.) ■
This integral is asking for the exact area under the curve of f(x) = 2 x −
Section 10.5: Basic Integral Rules Just like we were able to come up with rules for derivatives to find the exact instantaneous rate of
change at any point on a curve, we can also come up with some basic rules for computing
integrals. In this section, we will introduce only the simplest of integration rules. Other rules for
integrals can get very complicated and we do not need them for this level of material. However,
it's nice to see that there is a way to do all of this without having to use complicated computations
in a spreadsheet.
First, we need a definition of Antiderivatives:
Definition: Antiderivative
Given a function f(x), an antiderivative of f(x) is any function, F (x) , whose derivative is f(x).
That is:
If F ' ( x) = f ( x) , then F (x) is an antiderivative of f(x).
Chapter 10 – Page 204
Please note that we are making a distinction between f(x) and F (x) . They are not the same and
the difference in case is designed to highlight that fact.
For example, F ( x) =
x3
is an antiderivative of f ( x) = x 2 because:
3
3x 2
F ' ( x) =
= x 2 = f ( x) .
3
Notice that f(x) has many antiderivatives. F ( x) =
F ' ( x) =
x3
+ 5 also works, since:
3
3x 2
+ 5 = x 2 + 0 = x 2 = f ( x)
3
x3
and
3
get a valid antiderivative of f(x). To get around this multitude of antiderivatives, we'll simply
write the following for the antiderivative of f ( x) = x 2 , with c being any constant:
Because the derivative of any constant is 0, we can add any constant to the end of F ( x) =
F ( x) =
x3
+c
3
In this simple example you may have noticed that the antiderivative has a power that is one more
than the function in question. This makes total sense since the rule for taking the derivative of a
function of the form f ( x) = cx n involves reducing the power by 1. Finding antiderivatives is the
same as taking derivatives in reverse. Hence, the name antiderivative. With this in mind, we can
state the rule for finding antiderivatives of functions of the form f ( x) = cx n . As you will see, it's
the exact opposite of the rule for taking derivatives.
Rule for Antiderivatives
Given a function of the form f ( x) = cx n , one antiderivative of this function is:
F ( x) = c
x n +1
n +1
There are two things to notice about this rule. First, instead of subtracting 1 from the exponent,
we add 1 instead. Second, instead of multiplying c by the original power, we divide by 1 more
than the original power. As you can see, this is exactly the opposite of taking a derivative.
Chapter 10 – Page 205
Example 82 Find an antiderivative of f ( x) = x 4 + 2 x 3 + 5 x .
Solution:
We start by writing this as f ( x) = x 4 + 2 x 3 + 5 x1 so it's clear what the power on the 5 x
term is. Now we apply the antiderivative rule to each of these three terms.
x 4 +1
x 3 +1
x1+1
+2
+5
4 +1
3 +1
1+1
x5
x4
x2
=
+2 +5
5
4
2
5
4
2
5x
x
x
=
+
+
5
2
2
F ( x) =
Always simplify the fractions.
We can check this by taking the derivative of F (x ) and making sure it is equal to f(x).
5 x 4 4 x 3 10 x
+
+
5
2
2
4
3
= x + 2 x + 5x
F '( x) =
Answer: Since the check is complete, we can say that F ( x) =
x5 x 4 5x 2
+ +
+c.■
5
2
2
Example 83 Find an antiderivative for f ( x) = 0.25 x 3 + 3.5 x1.2 .
Solution:
Applying the rule, we get this:
x4
x 2.2
+ 3.5
4
2.2
= 0.0625 x 4 + 1.5909 x 2.2
F ( x) = 0.25
Answer: An antiderivative is F ( x) = 0.0625 x 4 + 1.5909 x 2.2 + c. ■
If we have a constant term in our function, say k, then the antiderivative of that term is kx,
where x is the variable that the function is defined by. For example, the antiderivative of 4 is 4x.
the antiderivative of 197.22734383 is 197.22734383q.
Chapter 10 – Page 206
Our next rule can come in handy but it’s just a special case of the rule above:
Antiderivative of Constants
Given f(x) = k, where k is a constant, an antiderivative of f(x) is F ( x) = kx .
Example 84 We know that the Demand function for our class project is given by:
D (q) = −0.00011519q 2 + 0.01501173q + 197.22734383
Find an antiderivative for D(q ) .
Solution:
This is a nice polynomial function so we can use our rule for antiderivatives.
Answer:
F (q) =
− 0.00011519q 3 0.01501173q 2
+
+ 197.22734383q .
3
2
Can you guess what we are going to use this for? ☺
Section 10.6: The Fundamental Theorem of Calculus With a title like the Fundamental Theorem of Calculus, you'd expect something grand from this
section, and that's exactly what we'll get. The Fundamental Theorem of Calculus gives us a way
to find the exact area under a curve that is well behaved (and all our functions in this course are
just that).
The easiest way to deal with the Fundamental Theorem of Calculus is to state it and then give
some examples.
The Fundamental Theorem of Calculus
If f(x) is a continuous function on the interval [a,b], then the area under f(x) from x = a to b is
the following:
b
Area =
∫ f ( x)dx = F (b) − F (a) ,
a
where F is an antiderivative of f(x).
Chapter 10 – Page 207
This important theorem tells us that if we want to compute the exact area under a curve, we first
find an antiderivative of the function and then evaluate the antiderivative at the endpoints of the
interval. When we subtract F (b) − F (a ) , we get the true area.
Example 85 Find the area under f ( x) = x 2 on the interval [2,4].
Solution:
We'll start with a picture of what we want:
y
20
10
x
0
1
2
3
4
We need to find
∫ x dx .
2
2
We start by finding an antiderivative of f(x):
F ( x) =
x 2 +1
x3
=
2 +1 3
Now we just evaluate F (b) − F (a) = F (4) − F ( 2) :
F (b) − F (a) = F (4) − F ( 2)
43 23
−
3
3
64 8
=
−
3 3
56
=
3
= 18.6666
=
Answer: The exact area is 56/3. ■
Chapter 10 – Page 208
4
5
Example 86 Find the area under f ( x) = x 3 + x 2 − 2 x + 4 on the interval [−2,2].
Solution:
Here's a picture of the area we're after:
y
10
5
x
-3
-2
2
We need to find
∫ (x
3
-1
0
1
2
3
+ x 2 − 2 x + 4 )dx .
−2
We start by finding an antiderivative of f(x):
x4 x3
x2
+
− 2 + 4x
4
3
2
4
3
x
x
=
+
− x 2 + 4x
4
3
F ( x) =
Now we just evaluate F (b) − F (a) = F (2) − F (−2) :
x4 x3
x2
+
− 2 + 4x
4
3
2
4
3
x
x
=
+
− x 2 + 4x
4
3
F ( x) =
F (b) − F (a) = F ( 2) − F (−2)
⎛ (2 )4 (2)3
⎞ ⎛ (− 2 )4 (− 2 )3
⎞
2
2
= −⎜⎜
+
− (2) + 4(2) ⎟⎟ − ⎜⎜
+
− (− 2) + 4(−2) ⎟⎟
3
3
⎝ 4
⎠ ⎝ 4
⎠
F ( 2)
Chapter 10 – Page 209
F (−2)
Make sure you subtract in the right order…we need to compute F (b) − F (a ) , not
F (a) − F (b) . Also, we need to take time with the arithmetic or we'll get an
incorrect result.
F (2) − F (−2) =
⎞
⎞ ⎛ (− 2)4 (− 2 )3
⎛ (2 )4 (2 )3
2
2
+
− (− 2) + 4(−2) ⎟⎟
= ⎜⎜
+
− (2 ) + 4(2) ⎟⎟ − ⎜⎜
3
3
⎠
⎠ ⎝ 4
⎝ 4
⎛ 16 8
⎞ ⎛ 16 − 8
⎞
= −⎜ + − 4 + 8 ⎟ − ⎜ +
− 4 − 8⎟
3
⎝ 4 3
⎠ ⎝ 4
⎠
⎛ 32 ⎞ ⎛ 32 ⎞ 32 32
= ⎜ ⎟ − ⎜− ⎟ =
+
3
⎝ 3 ⎠ ⎝ 3 ⎠ 3
64
=
3
= 21.3333
Answer: The exact area is 64/3. ■
Obviously, this is a very nice way to find exact areas that does not involve setting up a
spreadsheet. Of course, it does depend on knowing how to find the antiderivative, which is rarely
easy. And it can require a lot of messy calculations, as seen above.
Example 87 Find the area under the curve of f ( x) =
e −0.5 x
2
2π
between [−1,1].
Solution:
This is the formula for the standard normal distribution curve we've seen several times
before and here is the area we want.
y
0.4
0.2
x
-3
-2
-1
0
Chapter 10 – Page 210
1
2
3
We know that this area is about 0.68. The problem is that we don’t have a rule available
to us that gives us an antiderivative for f ( x) =
e −0.5 x
2
. This function is NOT of the
2π
form f ( x) = cx n , which is the only function we know to find antiderivatives for. Hence,
we are stuck! Our only alternative is to do what we already have. Namely, we have to
crank this out with an Excel spreadsheet. ■
We conclude this section's material by pointing out that with the Fundamental Theorem of
Calculus, we have united the two main areas of calculus. Derivatives and Integrals are the two
key components of calculus and, as we have seen, they are simply opposites of each other. More
formally, we would say that derivatives and integrals are inverses of each other, much like e x
and ln(x) are inverses, as are x 2 and x . It's an amazing discovery that unifies these two fields.
When it was first discovered by Newton and Leibniz in the 1600's, it completely revolutionized
mathematics. This new tool helped them solve long−standing, difficult problems that had
histories that went back for hundreds (even thousands) of years. That it can help us answer
questions like how to find Consumer's Surplus and probabilities is quite amazing.
Section 10.7: Integration with Graphmatica As you can imagine, Graphmatica can compute areas under curves. One example is probably
enough to illustrate this feature.12
Example 88 Find the area under the curve of f ( x) = x 2 + 4 x + 7 between x = −5 and x = 2 .
Solution:
Here is a graph of the function:
12
Ask your instructor for more help if you need it.
Chapter 10 – Page 211
To calculate the area under the curve, look for the “Integrate” command under the
Calculus menu:
When you use this command, a dialog box will appear which allows you to specify the
limits of integration. We enter -5 and 2 as follows and then click on “Caclulate to get
the Result:
As you can see above, the approximate area is 51.34143 square units.
Chapter 10 – Page 212
Clicking the “Calculate” button also produces a visual picture with the area shaded.
(This is how many of the graphs in this textbook were created.)
Note that Graphmatica does not find the antiderivative (symbolically) for you. It only
gives you a (pretty good) numerical estimate for the area under the curve.
Chapter 10 – Page 213
Section 10.8: Focus on the Project So how does integration help us with our class project?
First, remember that we know our Demand Function to be:
D (q ) = −0.00011519 q 2 + 0.01501173q + 197.22734383
We know from earlier in the chapter that the area under the Demand curve give us the total
possible revenue. Hence, we can compute this now. We want to find this area:
D(q)
200
100
q
0
500
1000
1500
To find the actual area, we need to know where the Demand curve crosses the q axis. To do this
we can use Excel's Solver tool to find when D( q) = 0 . Doing so gives us a result of about q =
1375.3
1,375.3 units. Hence, we want to compute
∫ D(q)dq . It's a bit messy, but it can be done.
0
1375.3
∫
0
1375.3
D (q) dq =
∫ (− 0.00011519q
2
+ 0.01501173q + 197.22734383)dq
0
1375.3
q3
q2
= − 0.00011519 + 0.01501173 + 197.22734383q
3
2
0
The calculations are
messy, but care in
Excel will easily do
the job.
= ...
≈ 185,562.05
This bar notation
means to evaluate at
1375.3 then subtract
what you get when
you evaluate at 0.
Chapter 10 – Page 214
Since D(q) is in dollars per drive and q is in thousands of drives, the value of the integral is in
thousands of dollars (i.e. 185,562.05 thousand dollars). BUT, Revenue is in millions of dollars, so
we divide out by 1,000.
185,562.05
= 185.56205
1,000
Thus, the total possible revenue is 185.56205 million dollars. ■
Next, recall the notion of Consumer's Surplus. Recall the following diagram:
Demand Function
D(q)
Consumer's Surplus
Revenue
Not Sold
q
The Consumer's Surplus is defined to be the area that is shaded in the picture. This, of course, is
not the area under the curve, but we can still use that concept to determine what this area is. First
of all, note that this area can be found by taking the following two areas and subtracting. When
we do so, what's left is the Consumer's Surplus
D(q)
D(q)
D(q0 )
q
q
q0
Chapter 10 – Page 215
Hence to find Consumer's Surplus, we will compute the following:
Consumer's Surplus =
q0
q0
0
0
= D(q )dq − D(q0 )dq
∫
∫
WHY are these two
equal to each other?
q0
∫
= D(q )dq − q0 D (q0 )
0
We'll need to keep in mind that q0 is the quantity of drives that maximizes our profit because the
project question asks for Consumer's Surplus at the production level that maximizes profit. Recall
that q0 = 575.644 (thousand) units maximizes profits.
At that level D(575.644) = 167.699 , so our total Revenue at this level of production is:
R (575.644) = (575.644)(167.699 )
= 96,534.75
Thus, since Revenue is in millions, our total revenue is 96,534.75 thousand dollars, or 96.535
million dollars. We can now compute our Consumer's Surplus:
q0
q0
0
0
Consumer's Surplus = D (q)dq − D(q0 )dq
∫
∫
q0
∫
= D (q)dq − 96.535
0
575.644
=
∫ (− 0.00011519q
2
+ 0.01501173q + 197.22734383)dq − 96.535
0
= 108,696 (thousands) − 96.535 (millions)
= 108.696 (millions) − 96.535 (millions)
= 12.161(millions)
The integral can be computed either by integrating "by hand" or using Excel. Either way, we get
108,695.8054 thousand dollars, or 108.696 million dollars. Thus, we have a total Consumer's
Surplus of 12.161 million dollars.
Chapter 10 – Page 216
Chapter 10 Practice Problems 1)
Consider the graph of a Demand function
p=D(q)
250
200
150
100
50
q
0
100
200
300
400
500
a.
b.
Use the graph to approximate the area of the shaded rectangle.
What does this area represent? Explain your result.
2)
Consider f ( x) = −2 x + 20 , whose graph is shown below:
600
y
20
15
10
5
x
-2
a.
b.
c.
d.
-1
0
1
2
3
4
5
6
7
8
9
10
11
Divide the interval [0,10] into five equal subintervals, each of width 2. (Thus, Δx = 2 .)
Sketch by hand the five rectangles whose heights are determined by the midpoint of each
subinterval.
Find the area of each rectangle.
Find the sum of the rectangular areas.
Is this sum a good approximation to the area under the graph of the function, f (x) , from x
= [0,10]? Why or why not?
Chapter 10 – Page 217
3)
Repeat the steps in the previous exercise for the function g ( x) = x 2 + 5 on the interval
[0,10]. The graph is shown below:
y
100
80
60
40
20
x
0
1
2
3
4
5
6
7
8
9
10
-20
4)
a.
b.
Let g ( x) = x 2 + 5 . To utilize Excel to find the approximate area above the x axis and
below the graph of g (x) , we first specify that we will use 500 subintervals. We keep the
interval [0,10].
Find the width of each subinterval.
Find the values that would go into cells B3, B4, C3, and C4. (Note that in Excel, once you
have two of them, you can drag down to generate as many more as you need.)
A
1
2
3
4
c.
i
0
1
2
B
xi
0
C
mi
--
D
f(mi)
--
E
th
i area
--
What formula would you type into Cell D3 to find the height of the first rectangle.
=
d.
What formula would you type into Cell E3 to find the area of the first rectangle.
=
e.
Find the total area under this curve on the interval [0,10]?
For Problems (5) to (8) find an antiderivative of the given function. Simplify your result as much
as possible.
5)
f ( x) = x + 3
6)
f ( x) = 3 x3 − 4 x 2 − 6 x + 1
Chapter 10 – Page 218
2 x 2 x3
+ +5
3
4
7)
f ( x) =
8)
f ( x) = 2 x 0.4 − 3x 0.6 + 2
For Problems (9) to (14), use the Fundamental Theorem of Calculus to evaluate the given
integral.
3
9)
∫ (x + 3)dx
0
10
10)
∫ 5dx
−10
8
11)
∫ (2 x
2
− 3)dx
2
− 3)dx
4
− 6 x 2 + x )dx
3
4
12)
∫ (2 x
1
3
13)
∫ (5 x
−2
2.5
14)
∫ (0.5 x
2
+ 0.4 x 3 )dx
0
15) The graph of f ( x) = − x 3 + x + 6 is shown. Find the area of the shaded region by using the
Fundamental Theorem of Calculus.
30
y
20
10
x
-2
0
2
-10
Chapter 10 – Page 219
4
16) The graph of f ( x) = 2 x 2 + x − 5 is shown. Find the area of the shaded region by using the
Fundamental Theorem of Calculus.
y
150
100
50
x
-4
-2
0
2
4
6
8
10
-50
17)) The graph of f ( x) = x 4 − x 3 − 6 x 2 + 30 is shown. Find the area of the shaded region by
using the Fundamental Theorem of Calculus.
y
100
50
x
-2
0
2
4
18) Find the area under the curve of f ( x) = −3x + 30 on the interval [4,8].
19) Find the area under the curve of f ( x) = − x 2 + 3 x + 5 on the interval [2,3].
20) Find the area under the curve of f ( x) = x 3 − 7 x + 2 on the interval [-2,2].
Chapter 10 – Page 220
21) The Demand function for a product is given by D(q ) = 40 − .02q . Use the Fundamental
Theorem of Calculus to find the following for the value q = 1500. Show all work:
a.
The Revenue Received.
b.
The Consumer’s Surplus.
c.
The Not Sold amount.
d.
Sketch a graph of the demand function and label (with numbers) each of the three regions
above.
1
q 2 . Use the
20000
Fundamental Theorem of Calculus to find the following for the value q = 300. Show all
work:
The Revenue Received.
The Consumer’s Surplus.
The Not Sold amount.
Sketch a graph of the demand function and label (with numbers) each of the three regions
above.
22) The Demand function for a product is given by D( q) = 20 −
a.
b.
c.
d.
For Problems (23) to (29), use Graphmatica to evaluate the given integral or find the requested
quantity. Copy the resulting graph and write down the resulting area as your final answer.
3
23)
∫ (x + 3)dx
0
10
24)
∫ 5dx
−10
8
25)
∫ (2 x
2
− 3)dx
2
− 3)dx
4
− 6 x 2 + x )dx
3
4
26)
∫ (2 x
1
3
27)
∫ (5 x
−2
2.5
28)
∫ (0.5 x
2
+ 0.4 x 3 )dx
0
Chapter 10 – Page 221
6
29)
⎛
x2 −1⎞
⎜⎜ 5 + 2
⎟dx
x + 1 ⎟⎠
2⎝
∫
30) The graph of f ( x) = − x 3 + x + 6 is shown. Find the area of the shaded region using the
Fundamental Theorem of Calculus and then check it with Graphmatica.
y
30
20
10
x
-2
0
2
4
-10
31) The graph of f ( x) = 2 x 2 + x − 5 is shown. Find the area of the shaded region using the
Fundamental Theorem of Calculus and then check it with Graphmatica.
150
y
100
50
x
-4
-2
0
2
4
6
-50
Chapter 10 – Page 222
8
10
32)) The graph of f ( x) = x 4 − x 3 − 6 x 2 + 30 is shown. Find the area of the shaded region using
the Fundamental Theorem of Calculus and then check it with Graphmatica.
y
100
50
x
-2
0
2
4
33) Find the area under the curve of f ( x) = −3x + 30 on the interval [4,8]. Use the
Fundamental Theorem of Calculus and then check the result with Graphmatica.
34) Find the area under the curve of f ( x) = − x 2 + 3 x + 5 on the interval [2,3]. Use the
Fundamental Theorem of Calculus and then check the result with Graphmatica.
35) Find the area under the curve of f ( x) = − x 2 + 3x + 5 on the interval [-2,2]. Use the
Fundamental Theorem of Calculus and then check the result with Graphmatica.
Chapter 10 – Page 223
Answers to Selected Problems 1)
a.
60,000 square units
2)
3)
4)
5)
x2
+ 3x
2
6)
7)
x 4 2 x3
+
+ 5x
16
9
8)
9)
27/2
10)
11) 925/3
12)
13) 415/2
14)
15) 24
16)
17) 132.8
18)
19) 6.166
20)
21)
Chapter 10 – Page 224
a.
b.
c.
$15,000
$22,500
$2,500
22)
23) 27/2
24)
25) 925/3
26)
27) 415/2
28)
29) 23.402
30) 24
31) 302.67
32) 124.86
33) 48
34) 6.166
35) 8
Chapter 10 – Page 225
Index A
Addition Rule .............................................................................................................................. 148
Antiderivatives ............................................................................................................................ 204
Area under a curve ...................................................................................................................... 189
Average Rate of Change.............................................................................................................. 106
B
Break−Even point.......................................................................................................................... 97
C
Constant Rule .............................................................................................................................. 142
Constraint .................................................................................................................................... 180
Consumer Surplus ....................................................................................................................... 191
Cost function ................................................................................................................................... 5
Cost function, defined ................................................................................................................... 27
Critical point................................................................................................................................ 158
Critical points, classifying ........................................................................................................... 163
D
Demand function ............................................................................................................................. 5
Demand function, defined ............................................................................................................. 64
Derivatives................................................................................................................................... 133
Difference quotient...................................................................................................................... 121
E
Economies of scale .......................................................................................................................... 7
F
First Derivative Test .................................................................................................................... 160
Fixed costs..................................................................................................................................... 26
Forward Method .......................................................................................................................... 114
Forward-Backwards Method ....................................................................................................... 117
Fundamental Theorem of Calculus ............................................................................................. 207
I
Instantaneous Rate of Change ..................................................................................................... 111
Integrals ....................................................................................................................................... 203
L
Limit ............................................................................................................................................ 100
Local maximum........................................................................................................................... 159
Local minimum ........................................................................................................................... 159
Index – Page 226
M
Marginal analysis ........................................................................................................................ 114
Marginal change .......................................................................................................................... 114
Marginal cost............................................................................................................................... 114
Marginal profit ............................................................................................................................ 114
Marginal revenue......................................................................................................................... 114
Midpoint Rectangles.................................................................................................................... 200
Monopoly ........................................................................................................................................ 5
N
Natural monopolies ......................................................................................................................... 5
NORMDIST ................................................................................................................................ 189
NORMSDIST.............................................................................................................................. 189
Not Sold....................................................................................................................................... 193
Numerical Derivatives................................................................................................................. 135
O
Optimization................................................................................................................................ 158
Overhead costs .............................................................................................................................. 26
P
Perfect competitor ........................................................................................................................... 5
Piecewise function......................................................................................................................... 23
Power Rule .................................................................................................................................. 144
Profit Functions ............................................................................................................................. 84
R
Relative maximum ...................................................................................................................... 159
Relative minimum ....................................................................................................................... 159
Revenue function............................................................................................................................. 5
Revenue function, defined............................................................................................................. 78
S
Secant Line .................................................................................................................................. 106
Solver .......................................................................................................................................... 177
Stationary inflection points.......................................................................................................... 159
T
Tangent Line................................................................................................................................ 111
Total possible revenue................................................................................................................. 191
Trendlines...................................................................................................................................... 49
V
Variable costs ................................................................................................................................ 27
Index – Page 227