Optimization Strategies for Maximum and Minimum Problems

1. Chapter 4GraphingandOptimization Section 6 Optimization

2. Optimization Optimization problems are problems that involve finding the absolute maximum or minimum of functions. Common optimization problems involve finding area and perimeter, maximizing revenue and profit, and inventory control.

3. Example 1 Area and Perimeter What are the dimensions of a rectangle with area 225 square meters that has the least perimeter? Solution If L and W represent the length and width of the rectangle, respectively, then area A = L·W = 225 and perimeter P = 2·L + 2·W

4. Example 1 Area and Perimeter continued To maximize P, we differentiate and use techniques for finding the minimum. W = 0, W = –15, and W = 15 are partition numbers of P ´. Of these, only W = 15 is in the domain of the perimeter, P. W = 15 is the only critical number for P.

5. Example 1 Area and Perimeter continued We test the critical number W = 15using the second derivative test for concavity to determine if this gives a maximum, minimum, or neither. The least perimeter is when W = 15. A square of side 15 meters is the rectangle having least perimeter (60 m) with area 225 square meters.

6. Example 2 Maximizing Area A homeowner has $320 to spend on building a fence around a rectangular garden. • As shown in the drawing, three sides of the fence will be constructed with wire fencing at a cost of $2 per linear foot. The fourth side will be constructed with wood fencing at a cost of $6 per linear foot. • Find the dimensions and the area of the largest garden that can be enclosed with $320 of fencing.

7. Example 2 Maximizing Area continued A homeowner has $320 to spend on building a fence around a rectangular garden. • Solution Let x and y represent the dimensions of the garden. • The area of the garden is A = xy. • The cost of the fencing is • C = 2y + 2x + 2y + 6x = 8x + 4y. • The homeowner has $320 to spend for fencing so • 8x + 4y = 320 • 4y = 320 – 8x • y = 80 – 2x

8. Example 2 Maximizing Area continued • Withy = 80 – 2x and A = x·y, • then A = x(80 – 2x) = 80x – 2x2 Because x and y represent dimensions of a rectangle, both x and y must be nonnegative, x ≥ 0 and y ≤ 0. Since y = 80 – 2x then 80 – 2x ≥ 0 and 80 ≥ 2x so x ≤ 40. We are to maximize the area A(x) = 80x – 2x2 for 0 ≤ x ≤ 40. Find critical numbers of A. A´(x) = 80 – 4x = 0 80 = 4x so x = 20 Since A(x) is continuous on [0, 40], the absolute maximum, if it exists must occur at a critical number or an endpoint.

9. Example 2 Maximizing Area continued • Evaluate the area at the endpoints and the critical number to obtain A(0) = 0, A(20) = 800, A(40) = 0. • The largest of these values corresponds to an area of 800 square feet when x = 20. • With x = 20, and A = 800, it follows that y = 40. • The garden should be 20 feet by 40 feet with one 20-foot side of wood fencing.

10. Procedure Strategy for Solving Optimization Problems Step 1 Introduce variables, look for relationships among the variables, and construct a mathematical model of the form Maximize (or minimize) f(x) on the interval I. Step 2 Find the critical numbers of f(x). Step 3 Use the procedures developed in Section 4.5 to find the absolute maximum (or minimum) of f(x) on the interval I and the numbers x where this occurs. Step 4 Use the solution to the mathematical model to answer all the questions asked in the problem.

11. Example 3 Minimizing Perimeter Suppose the homeowner in Example 2 decides that 800 square feet for their garden is too small and decides to increase the area to 1,250 square feet. What is the minimum cost of building a fence that will enclose a garden with an area of 1,250 square feet? What are the dimensions of this garden? Assume that the cost of fencing remains unchanged. Recall that one side of the garden was to have a wooden fence and the other three sides were to have wire fencing.

12. Example 3 Minimizing Perimeter continued Solution In Example 2, we found the cost to be C = 8x + 4y. This representation for cost does not change if the size of the garden changes. We are to minimize C = 8x + 4y subject to x·y = 1,250 Since x and y represent distances, x> 0 and y >0. Because x·y = 1,250, neither x nor y can be zero.

13. Example 3 Minimizing Perimeter continued

14. Example 3 Minimizing Perimeter continued The second-derivative is positive at the only critical number of C(x) (x = 25) for x > 0. C(x) has a local minimum at x and this local minimum is also an absolute minimum. The cost of the fencing is C(25) = $400 and with x·y = 1,250 and x = 25, then y = 50. The minimum cost for enclosing a 1,250-square-foot garden is $400, and the dimensions are 25 feet by 50 feet, with one 25-foot side of wood fencing.

15. Example 4 Maximizing Revenue and Profit An office supply company sells x permanent markers each year at a price of $p per marker. The price-demand equation is p = 10 – 0.001x. What price should the company charge for the markers to maximize revenue? What is the maximum revenue? Solution Revenue = price ×demand R(x) = (10 – 0.001x)·x = 10x – 0.001x2 Both price and demand must be nonnegative so x ≥ 0 and p = 10 – 0.001x ≥ 0 so x ≤ 10,000

16. Example 4 Maximizing Revenue and Profit continued The mathematical model for this problem is Maximize R(x) = 10x – 0.001x2 for 0 ≤ x ≤ 10,000. Solve R´(x) = 10x – 0.002x = 0. x = 5,000 is a critical number. R″(x) = –0.002 < 0 for all x The revenue is a maximum when x = 5,000. R(5,000) = $25,000 For demand, x = 5,000, price is 10 = 0.001(5,000) = $5. The company achieves maximum revenue of $25,000 at a price of $5 per marker.

17. Example 5 Maximizing Profit Suppose the company in Example 4 has a cost of manufacturing markers of C(x) = 5,000 + 2x. What is the company’s maximum profit? What should the company charge for each marker, and how many markers should they produce? Solution Profit = Revenue – Cost P(x) = R(x) – C(x) = (10x – 0.001x2) – (5,000 + 2x) = 8x – 0.001x2 – 5,000 The mathematical model is Maximize P(x) = 8x – 0.001x2 – 5,000 for 0 ≤ x ≤ 10,000.

18. Example 5 Maximizing Profit continued P´(x) = 8 – 0.002x = 0 gives x = 4,000, the only critical number for P(x) on 0 ≤ x ≤ 10,000. P″(x) = –0.002 < 0 for all x. x = 4,000 gives maximum profit P(4,000) = $11,000. Using the price-demand equation (p = 10 – 0.001x) from Example 4, the price is p = 10 – 0.001(4,000) = $6. Maximum profit of $11,000 is realized when 4,000 markers are manufactured and sold for $6 each.

19. Example 6 Maximizing Profit Suppose that the government decides to tax the company in Example 5 at a rate of $2 for each marker they produce. With this additional cost, how many markers should they produce to maximize profit? What is the maximum profit? How much should they charge for the markers to attain maximum profit? Solution The tax of $2 per unit changes the cost equation. C(x) = original cost + tax = 5,000 + 2x + 2x = 5,000 + 4x

20. Example 6 Maximizing Profit The new profit function is P(x) = R(x) – C(x) = 10x – 0.001x2 – (5,000 + 4x) P(x) = 6x – 0.001x2 – 5,000 We maximize P(x) = 6x – 0.001x2 – 5,000 for 0 ≤ x ≤ 10,000 P´(x) = 6 – 0.002x = 0 gives the critical number x = 3,000. P″(x) = –0.002 < 0 for all x. Producing 3,000 items yields maximum profit. P(3,000) = $4,000 and price, p = 10 – 0.001(3,000) = $7

21. Example 6 Maximizing Profit The tax of $2 per marker increased the company cost by $2 per marker. The price that they should charge for the markers to obtain maximum profit is $7 (an increase of $1 per marker in cost to the consumer over the lower taxed model.) The company must absorb the other $1 of cost (and the resulting decrease in their profit from $11,000 to $4,000.) The $6,000 in taxes to the government cost the company $7,000 in profit and increased the total cost of markers sold to customers by $3,000.

22. Example 7 Maximizing Revenue A management training company finds that at a price of $400 per person, 1,000 people will attend their seminar. For each $5 decrease in the price, and additional 20 people will attend. How much should they charge in order to maximize revenue? What is the maximum revenue? Solution Let x represent the number of $5 price reductions. Then 400 – 5x represents the price per customer. 1,000 + 20x represents the number of customers. Revenue = (price per customer)(number of customers) = (400 – 5x)(1,000 + 20x)

23. Example 7 Maximizing Revenue continued Price cannot be negative so (400 – 5x) ≥ 0 so x ≤ 80 A negative value of x results in a price increase so x ≥ 0. The model for this example is Maximize R(x) = (400 – 5x)(1,000 + 20x) for 0 ≤ x ≤ 80. R(x) = 400,000 + 3,000x – 100x2 R´(x) = 3,000 – 200x = 0 gives x = 15 as a critical number. R(x) is continuous on [0, 80] so the absolute maximum occurs at a critical number or an endpoint. R(0) = 400,000; R(15) = 422,500; R(80) = 0 The price for attending the seminar when x = 15 is $325. The company should charge $325 for the seminar resulting in the maximum revenue of $422,500.

24. Example 8 Inventory Control A multimedia company anticipates that there will be a demand for 20,000 copies of a certain DVD during the next year. It costs the company $0.50 to store a DVD for one year. Each time it must make additional DVDs, it costs $200 to set up the equipment. How many DVDs should the company make during each production run to minimize its total storage and setup costs? Solution In this inventory control problem, a basic assumption is that demand is uniform. For example, if the year is broken down into days, the demand for each day is the same. Similarly for hours, minutes, seconds, etc.

25. Example 8 Inventory Control continued The company could decide to make all 20,000 DVDs at the beginning of the year, minimizing the setup cost. This would result in large storage costs. On the other hand, they could manufacture based on demand, minimizing the storage cost, but increasing the setup cost. Somewhere between these extremes is the optimal solution that minimizes total storage and setup costs. Let x represent the number of DVDs manufactured each time. Let y represent the number of production runs. The total set up cost is 200y. With uniform demand, the number of stored DVDs reduces from x to 0 between each production run with an average number stored each day of x/2.

26. Example 8 Inventory Control continued It costs $0.50 to store a DVD for one year so the total storage cost for a year is 0.5(x/2) = 0.25x. The total cost is total cost = setup cost + storage cost C = 200y + 0.25x

27. Example 8 Inventory Control continued

28. Example 8 Inventory Control continued The critical number x = 4,000 yields minimum cost and the minimum cost C(4,000) = $2,000. Since xy = 20,000, then y = 5. The company will minimize its total cost by making 4,000 DVDs five times during the year.