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Unformatted text preview: 2 = g(t) ≤ t ≤ 1 Then g'(t) = 14  58t = 0 for t = 7/29. Since g"(t)<0, t = 7/29 and x 1 = [22/29, 28/29] T . Here z = f(22/29, 28/29) = .69. ∇ f(x 1 ) = [40/29 10/29]. We find d 1 by solving max z = 40d1/29 + 10d2/29 st 3d1 + d2 ≤ 4 d1,d2 ≥ Optimal solution is d 1 = [4/3 0] T . Now x 2 = [22/29 28/29] T + t 1 [50/87 28/29] T = [22/29 + 50t/87 28/29  28t/29] where t 1 solves max f(22/29 + 50t/87, 28/29  28t/29) = .69+.46t  2.927t 2 = h(t). ≤ t ≤ 1 Then h'(t) = .46  5.854t = 0 for t = .0786. Since g"(t)<0, t 1 = .0786 and x 2 = [.804 .890] T For this point z = .708. Review Problem 13. We want to show that q maximizes g(p), where g(p) = q[1  (1p) 2 ] + (1  q)(1  p 2 ). Then g'(p) = 2q(1p) + (1q)(2p) = 0 for q = p. Since g''(p) = 2q  2(1q)<0, we know that choosing p = q will maximize g(p), as desired....
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This note was uploaded on 09/06/2009 for the course IEOR 160 taught by Professor Hochbaum during the Spring '07 term at Berkeley.
 Spring '07
 HOCHBAUM
 Operations Research

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