Consider the following LP: Maximize - 2√31 subject to I + 3 = 42 31 + 212 = 12 X1.X2 ≥ 0 (a) Express the problem in equation form: (b) Determine all the basic solutions of the problem; and classify them as feasible and infeasible. (c) Use direct substitution in the objective function to determine the optimum basic feasible solution. (d) Verify graphically that the solution obtained in (c) is the Optimum LP solution; hence, conclude that the optimum solution can be determined algebraically by considering the basic feasible solutions only. (e) Show how the infeasible basic solutions are represented on the graphical solution space.
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Solving each pair of equations simultaneously, we get the following basic solutions: x1 = 0, x2 = 14 x1 = 4, x2 = 12 x1 = 6, x2 = 10 x1 = 14, x2 = 0 x1 = 12, x2 = 4 x1 = 10, x2 = 6 Out of these, the first three solutions are feasible, while the last three are Show more…
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