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Max Z = 3x1 - x2
2x1 + x2 ≥ 2
x1 + 3x2 ≤ 3
x2 ≤ 4
& x1 ≥ 0, x2 ≥ 0
Max Z = 3x1 - x2 + 0s1 + 0s2 + 0s3 - M a1
2x1 + x2 - s1+ a1= 2
x1 + 3x2 + s2 = 3
x2 + s3 = 4
x1 , x2 , s1, s2, s3, a1 ≥ 0
As all Δj ≥ 0, optimal basic feasible solution is achieved. Hence the solution is Max Z = 9, x1 = 3, x2 = 0
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