•
Optimization problems are problems that involve finding the absolute maximum value or the absolute minimum value of a function.
•
We will be able to calculate:
–
Area and perimeter
– Revenue and profit
–
Inventory control.
Optimization Strategies
1.
Introduce variables, look for relationships among these variables, and construct a math model of the form: Maximize (minimize) f ( x ) on the interval I .
2.
Find the critical values of f’
( x ).
3.
Find the maximum (minimum) value of f ( x ) on the interval I .
4.
Use the solution to the mathematical model to answer all the questions asked in the problem.
Profit = Revenue - Cost
Note: Maximum revenue and maximum profit occur at 2 different points. The profit is maximum when
P’(x) = R’(x) – C’(x) = 0, so R’(x) = C’(x) means when the marginal revenue is equal to the marginal cost.

Problem 1
Find two numbers whose difference is 15 and whose product is a minimum.
Let x and y be the two numbers x – y = 15 -------------- x = 15 + y xy is a minimum
Set M = xy
M = (15+y)y = y 2 + 15y
Find critical values:
M’ = 2y + 15 = 0
y = -15/2 = -7.5
Check to see if it is a min or max
M’’ = 2 > 0
Concave up so yes we have a min
Therefore the numbers are 7.5 and -7.5 (15 + -7.5)
Problem 2
Find two positive numbers whose product is 21 and whose sum is a minimum.
Let x and y are two positive numbers xy = 21 ----------------------

x = 21/y x + y is a minimum
Set M = x+ y
M = 21/y + y = y + 21 y -1
Find critical values
M’ = 1 – 21/ y 2 = 0
– 21/ y 2 = -1
y 2 = 21 so y = 
21
the number is positive so y = 21
x =
21 y

21
21

21
Therefore the numbers are 21 and 21
Check to make sure it is a min
M’’ > 0 concave up, so yes

Problem 3
Find the dimensions of a rectangle with an area of 108 square feet that has the minimum perimeter.
Let x and y be the dimensions of the rectangle.
Area A = xy = 108  y = 108/x
Minimum perimeter P = 2x + 2y
P = 2x + 2 (108/x)
P = 2x + 216 x -1
P = 2 – (216/x 2 )= 0
- 216/x 2 = -2 x 2 = 108
x = 
108
the dimensions have to be positive so y = 108 = 6 3
x =
108 y

108
6 3

6 3
Therefore the numbers are 6 3 and 6 3
*Check to make sure it is a min
P’’ > 0 concave up, so yes
Problem 4
A company manufactures and sells x ePhones 10S (these are better than iPhones) per week. The weekly cost and price-demand equations are
C(x) = 30,000 + 110x p = 700 – 0.2x (price-demand function)
A) What price should the company charge and how many phones should be produced to maximize the weekly revenue? What is the maximum weekly revenue?
R = px = (700 – 0.2x)x = 700x – 0.2x
2
R’ = 700 – 0.4x = 0
x = 1750
R’’ < 0 so it is a max
Therefore the company should charge p = 700 – 0.2 (1750) = $350
And the maximum revenue is R (1750) = $612,500
B) Find the maximum profit, the production level that will realize the max profit, and the price the company should charge for each ePhone.

P = R – C (Profit function)
P = 700x - 0.2x
2 – (30,000 + 110x)
= 700x – 0.2x
2 - 30,000 - 110x
= -0.2x
2 + 590x – 30,000
P’ = -0.4x + 590 = 0
X = 1470
P’’ < 0 so it is a maximum
Price should charge is p = 700 – 0.2 (1470) = $405
Max profit is P(1475) = $405,125
Problem 5
A deli sells 640 sandwiches per day at a price of $8 each.
A) A market survey shows that for every $0.20 reduction in the original $8 price, 15 more sandwiches will be sold. How much should the deli charge for a sandwich in order to maximum revenue?
Let x be the number of $0.20 reduction in price
Then the number of sandwiches sold will be 640 + 15x
The price per sandwich is 8 – 0.20x
* Find the domain
The price cannot be negative so p ≥ 0
8 – 0.20x ≥ 0
X ≤ 40
So the domain is 0≤ x ≤ 40
*The revenue function is
R(x) = (640 + 15x)( 8 – 0.20x)
R(x) = 5120 – 128x + 120x – 3x 2
= -3x 2 – 8x + 5120
R’(x) = -6x – 8 = 0
X = -8/6 = -4/3 this is not a critical value because it is not in the domain
So the price should stay the same $8
B) What is the maximum revenue?
R = 640(8) = $5120

Problem 6
A car rental agency rents 200 cars per day at a rate of $30 per day. For each $1 increase in rate, 5 fewer cars are rented. At what rate should the cars be rented to produce the maximum income? What is the maximum income?
Let x be the number of $1 increase in rate
So the number of cars is 200 – 5x
And the rate is p = 30 + 1x
* Find the domain
The rate can not be negative
30 – x ≥ 0
x ≥ -30
So the domain is x ≥ 0
The revenue function is
R(x) = (200 – 5x) (30 + 1x)
= 6000 + 200x – 150x – 5x 2
= 5x 2 + 50x + 6000
R’= -10x + 50 = 0
-10x = -50
X = 5 is a critical value
Rate p = 30 + 1(5) = $35
The maximum income is R(5) = -5(5 2 ) + 50 (5) + 6000 = $6125