Question 11 - NCERT Exemplar - MCQs - Chapter 12 Class 12 Linear Programming (Term 1)
Last updated at Nov. 20, 2021 by Teachoo
NCERT Exemplar - MCQs
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NCERT Exemplar - MCQs
Last updated at Nov. 20, 2021 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 = π/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 β p = q 2p = q p = π/π β΄ Value of Z will be maximum if p = π/π So, the correct answer is (B)