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Unformatted text preview: and z” be positive? Justify this. Otherwise, if you disagree with the solutions manual, please justify why you disagree - a counter-example perhaps. You will receive full credit too if you just copy the answer below. However, I would rather you think. The solution from the textbook is: max s.t z = z + z 4x1 + x2 2x1 − x2 2x1 − 3x2 ≤4 ≤ 0. 5 = z − z All variables non-negative First note that in any basic feasible solution, z and z cannot both be positive. Then observe that if 2x1 − 3x2 > 0, then z = 0 and the objective function will equal z + z = z = 2x1 − 3x2 = |2x1 − 3x2 | while if 2x1 − 3x2 &l...
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- Fall '08