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3.4
FURTHER APPLICATIONS OF OPTIMIZATION
Copyright © Cengage Learning. All rights reserved.

2. Introduction
In this section we continue to solve optimization problems.
We will see how to maximize a company’s profit if we are not given the price function, provided that we are given information describing how price changes will affect sales.
We will also see that sometimes x should be chosen as something other than the quantity sold.

3. Example 1 – FINDING PRICE AND QUANTITY FUNCTIONS
A store can sell 20 bicycles per week at a price of $400 each. The manager estimates that for each $10 price reduction she can sell two more bicycles per week. The bicycles cost the store $200 each. If x stands for the number of $10 price reductions, express the price p and the quantity q as functions of x.
Solution:
Let
x = the number of $10 price reductions
For example, x = 4 means that the price is reduced by $40 (four $10 price reductions).

4. Example 1 – Solution
cont’d
Therefore, in general, if there are x $10 price reductions from the original $400 price, then the price p(x) is
The quantity sold q(x) will be

5. Example 2 – MAXIMIZING PROFIT (Continuation of Example 1)
Using the information in Example 1, find the price of the bicycles and the quantity that maximize profit. Also find the maximum profit.
Solution:
In Example 1 we found
p(x) = 400 – 10x Price
q(x) = x Quantity sold at that price

6. Example 2 – Solution Revenue is price times quantity, p(x) · q(x):
cont’d
Revenue is price times quantity, p(x) · q(x):
R(x) = (400 – 10x)(20 + 2x) p(x)q(x)
= x – 20x Multiplying out and simplifying
The cost function is unit cost times quantity:

7. Example 2 – Solution Profit is revenue minus cost:
cont’d
Profit is revenue minus cost:
We maximize profit by setting the derivative equal to zero:
200 – 40x = Differentiating P = x – 20x2

8. Example 2 – Solution The critical number is x = 5.
cont’d
The critical number is x = 5.
The second derivative, P"(x) = – 40, shows that the profit is maximized at x = 5.
Since x = 5 is the number of $10 price reductions, the original price of $400 should be lowered by $ ($10 five times), from $400 to $350.
The quantity sold is found from the quantity function:
q(5) = · 5 = q(x) = x at x = 5

9. Example 2 – Solution Finally, we state the answer clearly.
cont’d
Finally, we state the answer clearly.
Sell the bicycles for $350 each.
Quantity sold: 30 per week.
Maximum profit: $ From P(x) = x – 20x2 at x = 5
Although we did not need to graph the profit function, the diagram shown below does verify that the maximum occurs at x = 5.

10. Choosing Variables
In example 2 we did not choose x to be the quantity sold, but instead to be the number of $10 price reductions.
We chose this x because from it we could easily calculate both the new price and the new quantity.
Other choices for x are also possible, but in situations where a price change will make one quantity rise and another fall, it is often easiest to choose x to be the number of such changes.

11. Example 3 – MAXIMIZING HARVEST SIZE
An orange grower finds that if he plants 80 orange trees per acre, each tree will yield 60 bushels of oranges. He estimates that for each additional tree that he plants per acre, the yield of each tree will decrease by 2 bushels. How many trees should he plant per acre to maximize his harvest?
Solution:
We take x equal to the number of “changes”—that is, let
x = the number of added trees per acre

12. Example 3 – Solution With x extra trees per acre,
cont’d
With x extra trees per acre,
Trees per acre: x Original 80 plus x more
Yield per tree: – 2x Original yield less 2 per extra tree
Therefore, the total yield per acre will be
We maximize this by setting the derivative equal to zero:
–100 – 4x = Differentiating Y = 4800 – 100x – 2x2
x = – Negative!

13. Example 3 – Solution
cont’d
The number of added trees is negative, meaning that the grower should plant 25 fewer trees per acre.
The second derivative, Y"(x) = –4,
shows that the yield is indeed
maximized at x = –25.
Therefore:
Plant 55 trees per acre – 25 = 55

14. Example 4 – MINIMIZING PACKAGE MATERIALS
A moving company wishes to design an open-top box with a square base whose volume is exactly 32 cubic feet. Find the dimensions of the box requiring the least amount of materials.
Solution:
The base is square, so we define
x = length of side of base
y = height

15. Example 4 – Solution A = x2 + 4xy
cont’d
The volume (length · width · height) is x · x · y or x2y, which (according to the problem) must equal 32 cubic feet:
x2y = 32
The box consists of a bottom (area
x2) and four sides (each of area xy).
Minimizing the amount of materials means minimizing the surface area of the bottom and four sides:
A = x2 + 4xy

16. Example 4 – Solution
cont’d
As usual, we must express this area in terms of just one variable, so we use the volume requirement to express y in terms of x:
The area function becomes

17. Example 4 – Solution We minimize this by finding the critical number:
cont’d
We minimize this by finding the critical number:

18. Example 4 – Solution The second derivative
cont’d
The second derivative
is positive at x = 4, so the area is minimized.
Therefore, the dimensions using the least materials are:
Base: 4 feet on each side Height from y = 32/x2 at x = 4
Height: 2 feet

19. Minimizing the Cost of Materials
How would the preceding problem have changed if the
material for the bottom of the box had been more costly
than the material for the sides?
If, for example, the material for the sides cost $2 per
square foot and the material for the base, needing greater
strength, cost $4 per square foot, then instead of simply
minimizing the surface area, we would minimize total cost:

20. Minimizing the Cost of Materials
Since the areas would be just as before, this cost would be
Cost = (x2)(4) + (4xy)(2) = 4x2 + 8xy
From here on we would proceed just as before, eliminating
the y (using the volume relationship x2y = 32) and then
setting the derivative equal to zero.

21. Maximizing Tax Revenue
Governments raise money by collecting taxes. If a sales tax or an import tax is too high, trade will be discouraged and tax revenues will fall.
If, the tax rate is too low, trade may flourish but tax revenues will again fall.
Economists often want to determine the tax rate that maximizes revenue for the government. To do this, they must first predict the relationship between the tax on an item and the total sales of the item.

22. Maximizing Tax Revenue
Suppose, for example, that the relationship between the tax rate t on an item and its total sales S is
If the tax rate is t = 0 (0%), then the total sales will be
If the tax rate is raised to t = (16%), then sales will be

23. Maximizing Tax Revenue
That is, raising the tax rate from 0% to 16% will discourage $8 million worth of sales.
The graph of S(t) as shows how total sales decrease as the tax rate increases.
With such information (which may be found from historical data), one can find the tax rate that maximizes revenue.

24. Example 5 – MAXIMIZING TAX REVENUE
Economists estimate that the relationship between the tax rate t on an item and the total sales S of that item (in millions of dollars) is
Find the tax rate that maximizes revenue to the government.
Solution:
The government’s revenue R is the tax rate t times the total sales :

25. Example 5 – Solution
cont’d
To maximize this function, we set its derivative equal to zero:

26. Example 5 – Solution This gives a tax rate of t = 9%.
cont’d
This gives a tax rate of t = 9%.
The second derivative,
is negative at t = 0.09, showing that the revenue is maximized.
Therefore:
A tax rate of 9% maximizes revenue for the government. See the graph as shown.