Question

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. $p=\frac{q}{2}$
• C. p = 3q
• D. p = q

Answer 1

Answer: (B)

$p=\frac{q}{2}$

Answer 2

The correct answer is (B) : $p=\frac{q}{2}$.