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The corner points of the feasible region determined by the following system of linear inequalities: 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q >0 . Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is.(A) p = q (B) P = 2q (C) p = 3q (D) q = 3p |
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Answer» Answer is (D) At (0, 0) Z = 0 At (5, 0) Z = 5p At (3, 4) Z = 3p +4q At (0, 5) Z = 5q ∴ maximum of Z 3p + 4q = 5q ∴ 3p = q |
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