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Solve by Steps for Two-Phase Method
Max Z = 5x1 + 8x2
Subject to
3x1 + 2x2 ≥ 3
x1 + 4x2 ≥ 4
x1 + x2 ≤ 5
& x1 ≥ 0, x2 ≥ 0
Answer
Standard LPP
3x1 + 2x2 - s1+ a1 = 3
x1 + 4x2 - s2+ a2 = 4
x1 + x2 + s3 = 5
x1 , x2 , s1, s2, s3, a1, a2 ≥ 0
Auxiliary LPP
Max Z* = 0x1 + 0x2 + 0s1 + 0s2 + 0s3 -1a1 -1a2
As all Δj ≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we move to phase II.
Phase II
As all Δj ≥ 0, optimal basic feasible solution is achieved. Thus the solution is Max Z = 40, x1 = 0, x2 = 5
Maximize Z =3x+4x subject to x1+x2 =3 2x1+3x2 =4 x1,x2 =0
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