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E160 Operations Research I
Fall 2008
Homework #7 Solution
Chapter 12
Section 12.8
2. We wish to maximize L
2/3
K
1/3
subject to 2L + K = 10. It is
easier to maximize ln L
2/3
K
1/3
= (2/3)ln L + (1/3)ln K
(this is a concave function so we know that Lagrange
multipliers will yield a maximum). Forming the Lagrangian LAG
we find that
LAG = (2/3)ln L + (1/3)ln K + λ(10  2L  K)
(1)
∂
LAG/
∂
L = 2/3L  2λ = 0 , (2)
∂
LAG/
∂
K = 1/3K  λ = 0
(3)
∂
LAG/
∂
λ = 10  2L  K = 0 From (1) L = 1/3λ.
From (2) K = 1/3λ and from (3) 2(1/3λ) + (1/3λ) = 10 or
λ = 1/10. Then L = K = 10/3.
4. L = 30x
1
1/2
+ 20x
2
1/2
+ λ(100  x
1
 x
2
)  x
1
x
2
(1)
∂
L/
∂
x
1
= 15x
1
1/2
 λ  1 = 0
(2)
∂
L/
∂
x
2
= 10x
2
1/2
 λ  1 = 0
(3)
∂
L/
∂
x
3
= 100  x
1
 x
2
= 0
From (1) and (2), x
1
= 225/(λ + 1)
2
and x
2
= 100/(λ + 1)
2
.
Then (3) yields 325/(λ + 1)
2
= 100 or (λ + 1)
2
= 3.25 and
λ = .80. Now we find that x
1
= 225/3.25 = 900/13 = 69.23 and
x
2
= 400/13 = 30.77. If an extra dollar can be spent on
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 Spring '08
 ADLER
 Operations Research

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