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B. Solve the following LP algebraically. а. Max Z= 150X+ 160Y subject to: X+ Y< 50 X> 20 Y> 15 Χ. Y>0 b. Min Z = 2x, + 3x2 + x3 %3D subject to: x, +x, > 10 x, +x, > 15 x, + x, + x, 2 20 X1, X2, X3 20

B. Solve the following LP algebraically. а. Max Z= 150X+ 160Y subject to: X+ Y< 50 X> 20 Y> 15 Χ. Y>0 b. Min Z = 2x, + 3x2 + x3 %3D subject to: x, +x, > 10 x, +x, > 15 x, + x, + x, 2 20 X1, X2, X3 20

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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8. Solve the following LP algebraically. Max Z = 150X+ 160Y f. Max Z= 650X +910Y а. subject to: subject to: 15 -Y<105 2 X+ Y< 50 4X+ X> 20 31 49 -Y< 90 10 Y> 15 5 Χ Υ>0 91 41 Y<100 10 b. Min Z = 2x, + 3x2 + x3 10 subject to: X, Y>0 x, +x, > 10 g. Max Z= 5x, + 2x, + 8x, x, + x, > 15 subject to: x, + x, + x, > 20 X,- 2x, + < 420 X1, X2, X3 20 2x, + 3x, – x,< 610 c. Min Z = 50x, + 40x2 6x, - x, + 3x, < 125 subject to: X1, X2, X3 2 0 2x, + x, 2 12 h. Min Z= 30x, + 40x, 2x, + 9x, > 36 subject to: 2x, + 3x, 2 24 X; + 3x, > 6 X1, X2 20 x, +x, > 4 x, + 2x, < 8 X1, X2 2 0 d. Min Z= 10x, + 30x, + 40x, + 20x4 subject to: x2 + x, > 20 Min Z= 20x; + 20x, i. x, + x, + x3 +x4 > 50 subject to: X1, X2, X3, X4NO 10x, + 30x, < 120 e. Max Z= 30x, + 40x2 30x, + 30x, > 130 subject to: 10x, – 10x, = 30 x, + 2x, < 12 X1, x, 2 0 -x, + 2x, < 8 2x, + x, 2 16 X1, x2 20
Transcribed Image Text:8. Solve the following LP algebraically. Max Z = 150X+ 160Y f. Max Z= 650X +910Y а. subject to: subject to: 15 -Y<105 2 X+ Y< 50 4X+ X> 20 31 49 -Y< 90 10 Y> 15 5 Χ Υ>0 91 41 Y<100 10 b. Min Z = 2x, + 3x2 + x3 10 subject to: X, Y>0 x, +x, > 10 g. Max Z= 5x, + 2x, + 8x, x, + x, > 15 subject to: x, + x, + x, > 20 X,- 2x, + < 420 X1, X2, X3 20 2x, + 3x, – x,< 610 c. Min Z = 50x, + 40x2 6x, - x, + 3x, < 125 subject to: X1, X2, X3 2 0 2x, + x, 2 12 h. Min Z= 30x, + 40x, 2x, + 9x, > 36 subject to: 2x, + 3x, 2 24 X; + 3x, > 6 X1, X2 20 x, +x, > 4 x, + 2x, < 8 X1, X2 2 0 d. Min Z= 10x, + 30x, + 40x, + 20x4 subject to: x2 + x, > 20 Min Z= 20x; + 20x, i. x, + x, + x3 +x4 > 50 subject to: X1, X2, X3, X4NO 10x, + 30x, < 120 e. Max Z= 30x, + 40x2 30x, + 30x, > 130 subject to: 10x, – 10x, = 30 x, + 2x, < 12 X1, x, 2 0 -x, + 2x, < 8 2x, + x, 2 16 X1, x2 20
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