chapter 2 - 6. For 7x1 + 10x2, slope = -7/10 For 6x1 + 4x2,...

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6. For 7 x 1 + 10 x 2 , slope = -7/10 For 6 x 1 + 4 x 2 , slope = -6/4 = -3/2 For z = -4 x 1 + 7 x 2 , slope = 4/7 80 40 40 80 x 2 x 1 20 60 100 20 60 0 100 -20 -40 -60 -80 -100 (c) (a) (b) 9. x 2 x 1 0 200 100 -100 -200 100 200 300 (150,225) (150,100) 10. 2 2 4 x 2 x 1 + 2 2 = 6 1 3 5 0 1 3 5 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 Optimal Solution x 1 = 12/7, x 2 = 15/7 6 5 1 3 15 4
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x 1 + 2 x 2 = 6 (1) 5 x 1 + 3 x 2 = 15 (2) (1) x 5 5 x 1 + 10 x 2 = 30 (3) (2) - (3) - 7 x 2 = -15 x 2 = 15/7 From (1), x 1 = 6 - 2(15/7) = 6 - 30/7 = 12/7 17. Max 5 x 1 + 2 x 2 + 8 x 3 + 0 s 1 + 0 s 2 + 0 s 3 s.t. 1 x 1 - 2 x 2 + 1 / 2 x 3 + 1 s 1 = 420 2 x 1 + 3 x 2 - 1 x 3 + 1 s 2 = 610 6 x 1 - 1 x 2 + 3 x 3 + 1 s 3 = 125 x 1 , x 2 , x 3 , s 1 , s 2 , s 3 0 21. a. Let F = number of tons of fuel additive S = number of tons of solvent base Max 40 F + 30 S s.t. 2/5 F + 1 / 2 S 200 Material 1 1 / 5 S 5 Material 2 3 / 5 F + 3 / 10 S 21 Material 3 F , S 0 b. . x 2 x 1 0 10 20 30 40 50 10 20 30 40 50 60 70 Feasible Region Tons of Fuel Additive Material 2 Optimal Solution (25,20) Material 3 Material 1 Tons of Solvent Base c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. S
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29. 6 4 2 2 4 x 2 x 1 0 6 Feasible Region x 1 = 3, x 2 = 1 Optimal Solution 3 x 1 + 4 x 2 = 13 Objective Function Value = 13 32. a. 0
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This note was uploaded on 04/20/2010 for the course IE ie200 taught by Professor . during the Spring '10 term at 카이스트, 한국과학기술원.

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chapter 2 - 6. For 7x1 + 10x2, slope = -7/10 For 6x1 + 4x2,...

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