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.
With a sales tax, profit for q>0 becomes f(q) = q(100 
4q)
2q  2q  50 and f'(q) = 96  8q = 0 for q = 12. f(12)>f(0)
so
q = 12 is optimal. Thus the sales tax slightly reduces
output.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 ADLER
 Operations Research, Optimization, local maximum

Click to edit the document details