E160 Operations Research I
Spring 2009
Homework #7 Solution
Chapter 12
Section 12.9
2. KT conditions yield
(1) 1  2x
1
λ
1
= 0 (x
1
constraint)
(2) 1 2x
2
λ
1
= 0 (x
2
constraint)
(3) λ
1
(1  x
1
2
 x
2
2
) = 0 , λ
1
≥
0
Trying λ
1
>0, we find from (1) and (2) that x
1
= 1/2λ
1
and
__
x
2
= 1/2λ
1
. Then (3) yields 1/2λ
1
2
= 1 or λ
1
= \/2 /2.
__
__
Finally we find that x
1
= \2/2 and x
2
= \/2/2. Since the
constraint is convex and the objective function is linear, we
know that we have found an optimal solution to the problem.
5. Let L
1
=Labor purchased at $15 per hour L
2
= Labor
purchased at $25 per hour. We wish to solve max z = 270(L
1
+
L
2
)
1/2
K
1/3
 15L
1
 25L
2
 5K
st L
1
≤
160, All variables nonnegative
Ignoring the nonnegativity restrictions, the KT conditions
yield
(1) 135(L
1
+ L
2
)
1/2
K
1/3
 15  λ
1
= 0 (L
1
constraint)
(2) 135(L
1
+ L
2
)
1/2
K
1/3
 25 = 0 (L
2
constraint)
(3) 90(L
1
+ L
2
)
1/2
K
2/3
 5 = 0 (K constraint)
(4) λ
1
(160  L
1
) = 0, λ
1
≥
0
We try λ
1
>0 (which by (4) yields L
1
= 160). From (2)
(L
1
+ L
2
)
1/2
= 5K
1/3
/27 or (L
1
+ L
2
)
1/2
= 27K
1/3
/5. From (3)
90(27)K
1/3
/5 = 5 or K = 918,330. From (2)
(L
1
+ L
2
)
1/2
= 27K
1/3
/5 = 524.88 so L
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 Spring '07
 HOCHBAUM
 Operations Research, Derivative, Optimization, Binary relation, Convex function

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