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

Since the minimum occurs at $ (3, 0) $ and $ (1, 1) $, we have $ z(3, 0) = z(1, 1) $. Calculate $ z $ at these points:

 

$$ z(3, 0) = 3p + 0q = 3p, \quad z(1, 1) = p + q. $$

Equating the two:

$$ 3p = p + q \implies 2p = q \implies p = \frac{q}{2}. $$

Additionally, the minimum must be less than at $ (0, 3) $. Calculate $ z(0, 3) $:

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

For the minimum to occur at $ (3, 0) $ and $ (1, 1) $, we require:

$$ 3p < 3q \implies p < q. $$

Thus, the condition is $ p = \frac{q}{2} $.

Answer 2

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:

\(3p = p + q\)

\(2p = q\)

\(p = \frac{q}{2}\)

Therefore, the condition on p and q is \(p = \frac{q}{2}\).