Question

The corner points of 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. The condition on p and q so that, 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

To find the condition on \( p \) and \( q \) such that the minimum of \( Z = px + qy \) occurs at the points (3, 0) and (1, 1), we first calculate \( Z \) at these corners:

For (3, 0): \(Z = 3p + 0 \cdot q = 3p\)

For (1, 1): \(Z = 1 \cdot p + 1 \cdot q = p + q\)

To ensure the same minimum value at both points, we equate these expressions:

\[3p = p + q\]

Solving the equation for \( p \):

\[3p - p = q\]

\[2p = q\]

Thus, the condition required is:

\[p = \frac{q}{2}\]

This ensures that the minimum value of \( Z \) occurs at both (3, 0) and (1, 1).