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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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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. The 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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Maximum value of Z occurs at (3,4) and (0,5). Therefore Z (3,4) = Z (5,0) $\to$3p + 4q = 5q $\to$3p = q, which is option (D)

Answer: (D) q = 3p

answered Apr 18, 2013 by balaji.thirumalai
 

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