# 22.pdf

## Mathematics 224 with Chaoqun Huang at Purdue University *

Sibo Z.

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Lesson 22: §7.3 Optimizing Functions of Two Variables - Application (II) Objective: Application of the Second Derivative Test. E1 A grocery store carries two brands of cat food, a local brand that it obtains at the cost of 30 cents per can and a well-known national brand it obtains at the cost of 40 cents per can. The grocer estimates that if the local brand is sold for x cents per can, then approximately 70−5x+4y cans pf the local brand and 80+6x−7y cans of the national brand will be sold each day. How should the grocer price each brand to maximize total daily profit from the sale of the cat food? Solution: Formulation of the problem We want to maximize the total profit function f(x,y) = (70−5x + 4y)·(x−30) + (80 + 6x−7y)·(y −40) which is f(x,y) = −5x2 + 10xy −20x−7y2 + 240y −5,300 Finding the critical points fx = −10x + 10y −20 = 0 fy = 10x−14y + 240 = 0 ⇒ (53,55) is the only critical point. Classification using second derivative test fxx = −10,fyy = −14,fxy = 10 ⇒ D := fxxfyy −(fxy)2 = 40 > 0 Therefore, f(x,y) is maximized at (53,55). square The following is an example of optimizing a function with THREE variables with a constraint. E2 Suppose you width to construct a rectangular box with a volume of 32 ft 2. Three different materials will be used in the construction. The material for the sides costs $ 1 per square foot, the material for the bottom costs $ 3 per square foot, and the material for the 1 top costs $ 5 per square foot. What are the dimensions of the least expensive such box? Solution: Let x,y,z are the length, width and height respectively. We want to minimize the cost function f(x,y,z) = 2yz + 2xz + 3xy + 5xy subject to the constraint that xyz = 32 Eliminating z, we just need to minimize the function f(x,y) = 64x + 64y + 8xy ⇒ x = 2,y = 2 is the only solution. and the cost is minimized when x = 2,y = 2,z = 8. square E3 A manufacturer is planning to sell a new product at the price of $210 per unit and estimates that if x thousand dollars is spent on development and y thousand dollars is spent on promotion, consumers will buy approximately 640yy + 3 + 216xx + 5 units of the product. If manufacturing costs for this product are $135 per unit, how much should the manufacturer on promotion and how on development to generate the largest possible profit from the sale of this product? [Hint: Profit = (number of units)(price per unit - cost per unit) - the total amount spent on development and promotion. ] Solution: We want to maximize the total profit function P(x,y) = parenleftbigg640y y + 3 + 216x x + 5 parenrightbigg (210−135)−x−y Finding the critical points Px = 216·75·5(x + 5)2 −1 = 0 Py = 640·75·3(y + 3)2 −1 = 0 ⇒ x ≈ 280,y = 376 is the unique solution. Using the second derivate test, one can check that P(x,y) is indeed maximized when x = 280,y = 376. square 2 Remark 1 In real application, if there is only one critical point, we usually do not perform the second derivative test. But it is always a good to check by doing second derivative test. 3

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