HW7_Sol - E160 Operations Research I Fall 2008 Homework #7...

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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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HW7_Sol - E160 Operations Research I Fall 2008 Homework #7...

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