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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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 Fall '09
 VLADIMIRLBOGINSKI

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