Lesson 6
Procedure
Examples
Section 1.4 (cont.): Functional Models
January 27th, 2014
Lesson 6
Procedure
Examples
In this lesson we are continuing our work with word problems.
We will encounter a couple new types of problems, but we will
use the same basic strategy to solve them. Let us recall our
strategy here.
Steps for solving a story problem
1
Read the problem carefully.
2
Identify and label the variables.
3
Draw a picture (if necessary).
4
Write down relationships with the variables.
5
Write down the main equation.
6
Solve.
Lesson 6
Procedure
Example
Examples
A cylindrical can is to have a capacity of 36π cubic inches. The
cost of the material used for the top and bottom of the can is 5
cents per square inch, and the cost of the material used for the
curved side is 3 cents per square inch. Express the cost of
constructing the can as a function of its radius.
Questions such as this one about the cost of constructing an
item will be common in this course. They are often very similar
to questions about volume and surface area.
Lesson 6
r = radius of the base, h = height of the cylinder
Procedure
Examples
h
r
It’s stated that the capacity (i.e. the volume) of the can is 36π
cubic inches, and so we have the relation
πr 2 h = 36π.
Lesson 6
Procedure
Examples
The cost of constructing the can is going to be the sum of the
costs of constructing each piece of it, in this case the top, the
bottom and the curved side. We are told the cost per square
inch of constructing each of these pieces. For example, the top
costs 5 cents per square inch to construct. Since square inches
is a measure of area, to get the total cost of constructing the
top we take 5 times the area of the top:
Cost of constructing the top: 5(πr 2 )
Similarly, the cost of constructing the bottom is 5(πr 2 ), and
the cost of constructing the curved side is 3(2πrh) = 6πrh.
Thus the total cost C of constructing the can is
C = 10πr 2 + 6πrh.
Lesson 6
Procedure
Examples
As in the problems we saw last time, we are not yet done
because this is not a function of r . It is a function of r and h.
We must use another equation to solve for h in terms of r and
then substitute.
πr 2 h = 36π ⇒ h =
36
r2
Therefore
2
C (r ) = 10πr + 6πr
36
r2
= 10πr 2 +
216π
.
r
Lesson 6
Example
Procedure
Examples
Sally’s company has been selling lamps at the price of $50 per
lamp, and at this price consumers have been buying 3,000
lamps a month. Sally wishes to lower the price and estimates
that for each $1 decrease in the price, 1,000 more lamps will be
sold each month. She can produce the lamps at a cost of $29
per lamp. Express Sally’s monthly profit as a function of the
price that the lamps are sold, draw the graph, and estimate the
optimal selling price.
This problem is significantly different from the previous
examples we have covered. Most notably, it is not geometric in
nature, and so we will not be drawing a picture to help us solve
it.
Lesson 6
Procedure
Examples
The problem asks for us to express the monthly profit as a
fuction of the price at which the lamps are sold, so let
x = price of lamps (in dollars). The monthly profit P is given
by the following formula:
P = (profit per lamp)(# of lamps sold per month)
So if we can write the profit per lamp and the number of lamps
sold per month in terms of x, we will be done.
The profit per lamp is easy. Since the lamps are sold at x
dollars and it takes $29 to make a lamp, the profit per lamp is:
x − 29
Lesson 6
Procedure
Examples
Calculating the number of lamps sold per month is more
involved. Let’s look at what we were told in the problem.
We know that when the lamps are priced at $50 each, 3,000
lamps are sold a month. We also know that for each $1
decrease in price, there is a 1,000 unit increase in sales. This
second bit of information establishes a linear relationship
between the price of the lamps, x, and the number of lamps
sold per month. As such, the number of lamps sold per month
will be given by an equation for a line.
Lesson 6
Procedure
Examples
Remember that to write down the equation of a line, we only
need to know two points on that line. Then we can compute
the slope of the line and use point-slope form for the equation.
Recall that if (x1 , y1 ) is a point on a line with slope m, then the
point-slope form of the line is given by the equation
y − y1 = m(x − x1 ).
The x-values of our line will be the price per lamp, and the
y -values will be the number of lamps sold per month. Thus
one point on our line is (50, 3000). Since a $1 decrease in price
leads to a 1,000 unit increase in sales, (49, 4000) is another
point on our line.
Lesson 6
Procedure
Now we can compute the slope of the line. Remember that for
two points (x1 , y1 ) and (x2 , y2 ), the slope of the line containing
1
those points is m = yx22 −y
−x1 . Hence our line has slope
Examples
m=
4000 − 3000
1000
=
= −1000.
49 − 50
−1
We can use any point on our line for point-slope form, so we
may use the piont (50, 3000). Then our equation is
y − 3000 = −1000(x − 50).
In order to plug this into our equation for P, we must first
solve for y . This gives us
y = −1000x + 53, 000.
Lesson 6
Therefore the montly profit is given by the function
Procedure
Examples
P(x) = (x − 29)(−1000x + 53, 000).
Notice that this is a (factored) quadratic equation whose
leading coefficient is negative. Therefore its graph is a parabola
that opens downward. From the above form we see that the
two zeros occur when x − 29 = 0 and when
−1000x + 53, 000 = 0. Thus the zeros are x = 29, 53.
Recall that the vertex of a parabola occurs directly between the
zeros of that parabola (if it has any). Thus the x-value of the
vertex is the average of the two zeros. Hence for this parabola,
= 82
the vertex has x-value 29+53
2
2 = 41.
Lesson 6
Procedure
In order to compute the y -value of the vertex, we simply plug
the x-value of the vertex into our function:
Examples
P(41) = (12)(12, 000) = 144, 000.
Therefore our parabola has vertex (41, 144, 000). Since we
know the vertex and the two zeros, we have enough
information to graph the parabola.
Finally, we were asked to estimate the optimal selling price, i.e.
the selling price that generates the most monthly profit. The
graph of our monthly profit function is a parabola opening
downward and so has its maximum at the vertex. Therefore the
optimal selling price is the x-value of the vertex, x = $41.
Lesson 6
Example
Procedure
Examples
Farmers can get $8 per bushel for their potatoes on July 1, and
after that, the price drops by 5 cents per bushel per day. On
July 1, a farmer has 140 bushels of potatoes in the field and
estimates that the crop is increasing at the rate of 1 bushel per
day. Express the farmer’s revenue from the sale of the potatoes
as a function of the time at which the crop is harvested, draw
the graph, and estimate when the farmer should harvest the
potatoes to maximize revenue.
This problem is very similar to the previous one, and so we will
apply the same method.
Lesson 6
Procedure
t = days after July 1 (note that ”time” is too vague of a
description here). Let R be the revenue from the potatoes.
Then we have the function
Examples
R = (# of bushels sold)(price per bushel).
Both the number of bushels sold and the price per bushel are
given by linear relationships with respect to t.
First we compute the price per bushel. For each day past July
1, the price per bushel drops by 5 cents.
Two points: (0, 8), (1, 7.95) (note that t = 0 is July 1)
Slope: m =
7.95−8
1−0
= −0.05
Point-slope form: y − 8 = −0.05(t − 0)
Slope-intercept form: y = −0.05t + 8
Lesson 6
Procedure
Examples
Now we compute the number of bushels sold. For each day
past July 1, 1 more bushel will be sold.
Two points: (0, 140), (1, 141)
Slope: m =
141−140
1−0
=1
Point-slope form: y − 140 = 1(t − 0)
Slope-intercept form: y = t + 140
Therefore the revenue from potatoes is given by the function
R(t) = (t + 140)(−0.05t + 8).
This is also the equation of a parabola that opens downward.
Zeros: t = −140, 160, Vertex: (10, 1125)
The farmer should harvest t = 10 days after July 1, i.e. on July
11, to maximize revenue.
Lesson 6
Procedure
Examples
Example
A company has received an order from the city recreation
department to manufacture 8,000 Styrofoam kickboards for its
summer swimming program. The company owns several
machines, each of which can produce 30 kickboards an hour.
The cost of setting up the machines to produce these particular
kickboards is $20 per machine. Once the machines have been
set up, the operation is fully automated and can be overseen by
a single production supervisor earning $19.20 per hour. Express
the cost of producing the 8,000 kickboards as a function of the
number of machines used, draw the graph, and estimate the
number of machines the company should use to minimize cost.
Like the first example, this relates to cost. But it is not
geometric, and the method for solving it will be significantly
different.
Lesson 6
Procedure
Examples
We are asked to write a function in terms of the number of
machines used. It is clear that the number of hours the
machines are running will also affect cost, so we make two
variables.
x = # of machines used, t = # of hours machines run
The cost of constructing the kickboards is given by
C = 20x + 19.20t.
The 20x is the cost from starting the necessary machines, and
the 19.20t is the cost from keeping those machines running for
the necessary amount of time.
Lesson 6
Again, this is not a function of x. We must solve for t in terms
of x and then substitute.
Procedure
Examples
We are told that each machine can produce 30 kickboards an
hour. Thus if x machines are used, the company will produce
30x kickboard per hour. If those x machines run for t hours,
then the company produces 30xt kickboards in total. We know
the company wants to produce 8000 kickboards, and so we
have the equation 30xt = 8000. From this we have t = 800
3x .
Therefore the cost of producing the 8000 kickboards is given by
the function
5120
800
C (x) = 20x + 19.20
= 20x +
3x
x
Lesson 6
Procedure
Examples
The graph of this function is more complicated than those in
the previous examples. It’s actually a hyperbola. Restricting
ourselves to the first quadrant (where both x and C are
positive), we see that the minimum cost occurs around x = 16.