HW_Solution_Chapter 11_EMSE102-202

HW_Solution_Chapter 11_EMSE102-202 - 1 OR CHAPTER 11...

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1 OR CHAPTER 11 SOLUTIONS SECTION 11.2 _____________________________ 2. max z= \/s(s - a)(s - b)(s - c) st a + b + c = 60 s = (a + b + c)/2 a b + c b a + c c a + b a 0, b 0, c 0 _____ ______ ______ 8 MAX =10*X1^.4+8*X2^.5+12*X3^.3+16*X4^.6; X1+X2+X3+X4=100000; X1>0; X2>0; X3>0; X4>0; SECTION 11.3 3. f''(x) = 2x -3 >0 (for x>0). Thus f(x) is convex on S. 4. f''(x) = a(a - 1)x a-2 0 so f(x) is concave on S. 7. f(x 1 , x 2 ) is the sum of two convex functions and is therefore a Convex function. ( Look at slide #56 of nonlinear programming )

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2 SECTION 11.5 1. 0.8 (0.618)^k<0.8 K*ln (0.618)<ln (0.1) K<ln(0.1)/ln(0.618) k>4.7844 k=5 iteration is needed Iteration 1 : a = -3 b = 5, so b - a = 8. x 1 = 5 - .618(8) = .056 x 2 = -3 + .618(8) = 1.944. f(x 1 ) = .115, f(x 2 ) = 7.67 f(x 2 )>f(x 1 ) so interval of uncertainty is now (.056,5]. Then Iteration 2 : X 3 = 5-0.618 (5-0.056) =1.944, x 4 = .056 + .618(4.944) = 3.11. f(x
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This note was uploaded on 01/02/2010 for the course EMSE 202 taught by Professor Mohaghegh,abeledo during the Spring '07 term at GWU.

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HW_Solution_Chapter 11_EMSE102-202 - 1 OR CHAPTER 11...

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