Q6.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
Answer : Option AExplaination / Solution: Here Z = px +qy , subject to constraints : 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 As it is given that Z is maximum at ( 3 ,4 ) and ( 0, 5 ). Therefore , 3p + 4q = 0p + 5q , which gives 3p = q .