# Lesson 12.6: Optimization ## Lesson 12.6: Optimization

• 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.

4.

Find the maximum (minimum) value of

f

(

x

) on the interval

I

. 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 = x = 21

y

 21 21  21 21 Therefore the numbers are 21 and Check to make sure it is a min M’’ > 0 concave up, so yes 21

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 = x = 108

y

 108 6 3  6 3 108 = 6 3 Therefore the numbers are 6 3 and 6 *Check to make sure it is a min P’’ > 0 concave up, so yes 3

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