HW3_Sol - E160 Operations Research I Fall 2008 Homework #3...

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E160 Operations Research I Fall 2008 Homework #3 Solution Chapter 12 Section 12.4 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. 2. Let f(q) = profit if q units are produced. Then f(0) = 0 and for q>0, f(q) = q(100 - 4q) - 2q - 50. For q>0 f'(q) = 98 - 8q = 0 for q = 12.25. Now f''(q) = -8 0 so q = 12.25 is a local maximum (and a maximum over all x>0). Now f(12.25) = 550.25>f(0) = 0, so 12.25 units should be produced.
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This note was uploaded on 12/18/2008 for the course IEOR 41033 taught by Professor Adler during the Spring '08 term at University of California, Berkeley.

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HW3_Sol - E160 Operations Research I Fall 2008 Homework #3...

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