Question 11 - NCERT Exemplar - MCQs - Chapter 12 Class 12 Linear Programming
Last updated at April 16, 2024 by Teachoo
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Last updated at April 16, 2024 by Teachoo
Question 11 Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is (A) p = 2q (B) p = 𝑞/2 (C) p = 3q (D) p = q Given corner points are (0, 3), (1, 1), (3, 0) Max. Z = px + qy Since maximum value of Z occurs on (3, 0) and (1, 1) Hence, Value on (3, 0) = Value on (1, 1) 3p = p + q 3p − p = q 2p = q p = 𝒒/𝟐 ∴ Value of Z will be maximum if p = 𝒒/𝟐 So, the correct answer is (B)