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

The Correct Option is B

The corner points are (0, 3), (1, 1), and (3, 0).

The objective function is z = px + qy, where p > 0 and q > 0.

The minimum of z occurs at (3, 0) and (1, 1).

At (3, 0), z = p(3) + q(0) = 3p.

At (1, 1), z = p(1) + q(1) = p + q.

Since the minimum occurs at both (3, 0) and (1, 1), we must have:

3p = p + q

2p = q

p = q/2

Therefore, the condition on p and q is p = q/2.