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the primal problem remains feasible. If A be increase in the first resource, it can be determined from the condition 2/5 -1/5715 +A f60 1/5 2/5 8+2A) 10/5+ 2A/5 - 2751 5 575+/5+415 9+A [1] [215 215) IV As x, and xz remain feasible (20) for all values of A 20, the first resource can be increased indefinitely while maintaining the condition that each additional unit will increase the objective function by 29/5 (e) The second resource should be decreased as each additional unit of the second resource decreases the objective function by 2/5. Let A be the decrease in the second resource. To find its extent, we make use of the condition that the current solution remains feasible so long as 275 -1/51 5 1/5 2152-A 8+ 10/5-2/5 +4/51 5 [5/5+4/5-24/5) 9-2A [1] 00 9-24 Evidently x remains positive only so long as 20 or A S 9/2 5 If A > 9/2, X, becomes negative and must leave the solution. EXAMPLE 6.6-1.2 Given the following L.P. problem: Maximize Z-X+ 2x-* subject to 3x, + xy - xy S 10 -X + 4x2 + xy26 x + xy S4 X, X, Xy 20, and the optimal solution given by table 6.59, find the separate ranges of b, b, and by consistent with the optimal solution. TABLE 6.59 0 0 -M Basis X X2 X3 X5 Az 0 -1 0 6 1 1 0 + 0 1 -1 10 2 2 0 2 0 -1 -3 0 -2 -M X4 X 0 X4 X2 3 0 2 0 1 4 3 0 0 1 0 0 2 0 0 0