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z = 20x1 + 32x2 + 40x3,
subject to
3x1 + x2 + 6x3 ≥ 9
x1 + x2 ≥ 9
4x2 + x3 ≥ 12
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

solve for x1, x2, x3 and z

1 Answer

5 votes

Answer:

Explanation:

1. constraints with "
\geq" we should subtract surplus variable S1, S2, S3 and add artificial variable A1, A2, A3

Hence Z = 20 x1 + 32x2 + 40x3 + 0S1 + 0S2 + 0S3 + MA1 + MA2 + MA3

subject to


3x_1 +x_2+6x_3-S_1+A_1=9


x_1+x_2-S_2+A_2=9


4x_2+x_3-S_3+A_3=12

and
x_1,x_2,x_3,S_1,S_2,S_3,A_1,A_2,A_3\geq 0

User Deann
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