ESI6314_practice_problems_Chapter12_solutions

ESI6314_practice_problems_Chapter12_solutions - ESI 6314...

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ESI 6314 Practice Problems for Chapter 12 Solutions Section 12.4: Problem 1 Let f(x) = profit if $x is spent on advertising. Then f(0) = 0 and for x>0, f(x) = 300x 1/2 - 100x 1/2 - 5000 - x. Since f(x) has no derivative at x = 0, maximum profit occurs either for x = 0 or a point where f'(x) = 0. Now for x>0 f'(x) = 100x -1/2 - 1 = 0 for x = 10,000. Also f''(x) = -50x -3/2 <0 for x>0. Thus x = 10,000 is a local maximum(and a maximum over all x>0). We now compare f(0) and f(10,000) to determine what the company should do. f(0) = 0 and f(10,000) = $5,000, so company should spend $10,000 on advertising. If fixed cost is $20,000, f'(x) = 0 still holds for x = 10,000. Comparing f(0) = 0 and f(10,000) = -10,000, we now find that x = 0 is optimal. Section 12.4: Problem15 Let n = number of warehouses. Then we wish to minimize f(n) = cost per year with n warehouses. f(n) = 100,000n + 160,000(100/n) 1/2 . f'(n) = 100,000 + 1,600,000 (-.5n
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ESI6314_practice_problems_Chapter12_solutions - ESI 6314...

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