Lesson 32
Procedure
Examples
Section 3.5: Additional Applied Optimization
April 4th, 2014
Lesson 32
Procedure
Examples
In this lesson, we will take a look at more applications of
optimization. Unlike the word problems in the last lesson, the
examples in this lesson will generally not provide us with the
function that we are supposed to be optimizing. We will have
to construct it ourselves based on the description of the
problem.
Lesson 32
Procedure
Examples
Guidelines for solving optimization problems
(1) Decide what is to be optimized and label variables.
(2) Express relationships between variables in terms of
equations or inequalities.
(3) Express the quantity to be optimized in terms of one
variable.
(4) Find the required minimum or maximum of the function.
Lesson 32
Procedure
Example
Find two positive numbers whose sum is 50 and whose product
is as large as possible.
Examples
We let x and y be the two positive numbers. We can assume
that each one is bigger than 0 and at most 50, i.e.
0 < x, y ≤ 50.
Since we know we want the sum to be 50, we get the equation
x + y = 50.
We are asked to make the product as large as possible, subject
to the stated constraint. Thus we want to maximize the
product fuction P = xy . Since this is a function of two
variables, we must write one variable in terms of the other
before proceeding.
Lesson 32
Procedure
Examples
Since x + y = 50, we have y = 50 − x. Therefore we can write
P(x) = x(50 − x) = 50x − x 2
We can now maximize this function.
P 0 (x) = 50 − 2x = 2(25 − x) ⇒ P 0 (x) = 0 at x = 25.
Since this is the only critical value in our interval i = (0, 50],
we can use our procedure from the previous lesson to see if this
is a maximum. Remember, this involves evaluating P 00 (x) at
the critical value and looking at the sign.
P 00 (x) = −2 ⇒ P 00 (25) = −2 < 0 (concave down)
Thus P(x) has a maximum at x = 25. When x = 25,
y = 50 − 25 = 25. Therefore the two numbers are (25, 25).
Lesson 32
Example
Procedure
Examples
A manufacturer can produce T-shirts at a cost of $3 apiece.
The shirts have been selling at $7 apiece, and at this price they
have been selling 3000 shirts a month. An increase of $1 in
selling price decreases sales by 300 shirts. At what price should
the manufacturer sell the shirts to maximize profit?
We want to maximize profit. Let x be the selling price of the
t-shirts. Recall that
Total profit = (profit per item)(# of items sold),
and the profit per item is the revenue of each item minus the
cost of each item. Let P be the total profit.
Lesson 32
P(x) = (x − 3)(mx + b)
Procedure
Examples
We use mx + b for the number of items sold because it is given
by a linear relationship (a $1 increase in price results in a 300
unit decrease in sales). Since 3000 shirts are sold per month
when the price is $7, one point on the line is (7, 3000).
Another is (8, 2700). Thus we have
Slope: m =
2700−3000
8−7
= −300
Line: y − 3000 = −300(x − 7) ⇒ y = −300x + 5100
Therefore
P(x) = (x − 3)(−300x + 5100) = −300(x − 17)(x − 3)
for 0 < x < ∞.
Lesson 32
We must find the maximum of P(x) = −300(x − 17)(x − 3).
Procedure
Examples
P 0 (x) = −300(x − 3) − 300(x − 17)
= −300x + 900 − 300x + 5100
= −600x + 6000
= −600(x − 10)
Therefore P 0 (x) = 0 ⇒ x = 10. To make sure this is a
maximum of P(x), we use the second derivative test.
P 00 (x) = −600
⇒
P 00 (10) = −600 < 0
Therefore x = 10 is the absolute maximum of P(x). Thus
profit is maximized when the shirts are sold for $10 apiece.
Lesson 32
Procedure
Examples
Example
Each machine at a certain factory can produce 50 units per
hour. The setup cost is $80 per machine, and the operating
cost is $5 per hour. How many machines should be used to
produce 8,000 units at the least possible cost? (Remember
that the answer should be a whole number.)
Here we want to minimize cost. Let x be the number of
machines used, and let t be the number of hours that the
factory operates. The total cost will depend on both of these
variables. In fact,
Total Cost = C = 80x + 5t,
0 < x, t < ∞
where 80x is the cost of setting up x machines and 5t is the
cost of operating the factory.
Lesson 32
Procedure
In t hours, x machines will produce 50xt units (since each
machine produces 50 units per hour). Since we are producing
8,000 units, we have the equation
Examples
50xt = 8000.
8000
50x
=
160
x .
Therefore
160
800
C (x) = 80x + 5
= 80x +
,
x
x
Thus t =
2
0<x <∞
Now C 0 (x) = 80 − 800
= 80(xx 2−10) , and so the critical number
√ x2
√
00 ( 10) > 0, this
of C (x) is x = 10. Since C 00 (x) = 1600
⇒
C
3
x
is a minimum. The cost for three machines is C (3) = 506.67,
and the cost for four machines is C (4) = 520. Thus the factory
should use three machines.
Lesson 32
Example
Procedure
Examples
Granville Thomas is a cirtrus grower in Florida. He estimates
that if 70 orange trees are planted, the average yield will be
350 oranges per tree. The average yield will decrease by 5
oranges per tree for each additional tree planted on the same
acreage. How many trees should Granville plant to maximize
the total yield?
Here we want to maximize total yield. Let x be the number of
trees planted, and let Y be the total yield. Of course we have
Total yield = (# of trees)(yield per tree)
= x(mx + b)
with 0 < x < ∞.
Lesson 32
The yield per tree is given by a linear relation. Two points on
the line are (70, 350) and (71, 345).
Procedure
Slope: m =
Examples
345−350
71−70
Line: y − 350 = −5(x − 70)
=
−5
1
⇒
= −5
y = −5x + 700
Therefore
Y (x) = x(−5x + 700) = −5x 2 + 700x.
Now Y 0 (x) = −10x + 700 = −10(x − 70), and so the critical
value of Y (x) is x = 70. Since
Y 00 (x) = −10 ⇒ Y 00 (70) = −10 < 0, this is the absolute
maximum of Y (x). Therefore Granville should plant 70 trees to
maximize yield.