Question
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
- p = q
- p = 2 q
- p = 3 q
- q = 3 p
Solution
D.
q = 3 pThe corner points are (0, 0), (5, 0), (3, 4), (0, 5)
At (0, 0), Z = 0 + 0 = 0
At (5, 0), Z = 5 p + 0 = 5 p
At (3, 4), Z = 3 p + 4 q
At (0, 5), Z = 0 + 5 q = 5 q
Since maximum value occurs at both (3, 4) and (0, 5).
∴ 3 p + 4 q = 5 q
∴ q = 3 p
∴ (D) is correct answer.
