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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.
 Spring '08
 ADLER
 Operations Research

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