Fall 2009
IEOR 160
Industrial Engineering & Operations Research
September 8, 2009
Page 1 of 3
HOMEWORK 1 SOLUTIONS
Chapter 12.1
1. lim
h
→
0
3
h
+
h
2
h
= lim
h
→
0
3 +
h
= 3.
3.
(a)

xe
−
x
+
e
−
x
(b)
2
x
(
x
2
+1)
−
2
x
3
(
x
2
+1)
2
(c) 3
e
3
x
(d)

6
(3
x
+2)
3
(e)
3
x
4.
∂f
∂x
1
= 2
x
1
e
x
2
∂f
∂x
2
=
x
2
1
e
x
2
∂
2
f
∂x
2
1
= 2
e
x
2
∂
2
f
∂x
1
∂x
2
=
∂
2
f
∂x
2
∂x
1
= 2
x
1
e
x
2
∂
2
f
∂x
2
2
=
x
2
1
e
x
2
5. ln(1 +
h
) =
h

h
2
2
+
2
p
3
3!
for
p
∈
[1
,
1 +
h
].
Chapter 12.2
1.
(a) Let
S
= soap opera ads and
F
= football ads. Then we wish to
min
z
= 50
S
+ 100
F
s.t. 5
√
S
+ 17
√
F
≥
40
20
√
S
+ 7
√
F
≥
60
S
≥
0
,F
≥
0
(b) Since doubling
S
does not double the contribution of
S
to each constraint, we are violating
the proportionality assumption. Additivity is not violated.
(c) This accounts for the fact that an extra soap opera ad yields a benefit which is a de
creasing function of the number of football ads. This accounts for the fact that we may
not want to double count people who see both types of ads.
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 Fall '07
 HOCHBAUM
 Operations Research, Industrial Engineering, Optimization, heating oil, Industrial Engineering & Operations Research

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