Ex 12.2, 11 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 (B) q = 3p Constraints 2x + y ≤ 10 x + 3y ≤ 15 x, y ≥ 0 Max. Z = px + qy Since maximum value of Z occurs on (3, 4) and (0, 5) Hence, Value on (3, 4) = Value on (0, 5) 3p + 4q = 5q 3p = 5q − 4q 3p = q ∴ Value of Z will be maximum if q = 3p ∴ (D) is correct answer

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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