Question

The corner points of the feasible region determined by a system of linear constraints are \((0,3),(1,1)\) and \((3,0)\). If the objective function is \(z = px+qy\), where \(p,q > 0\), then the condition on p and q such that the minimum of z occurs at both \((3,0)\) and \((1,1)\) is _____

• A. \(p = 3q\)
• B. \(3p = q\)
• C. \(p = \frac{q}{2}\)
• D. \(p = 2q\)

Hint:

If the minimum is at two corners, the objective function has equal value there.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When the minimum of \(z=px+qy\) occurs at two corner points, \(z\) has the same value at both (and at every point on the segment joining them).

Step 2: Equate z values:
At \((3,0)\): \(z=3p\). At \((1,1)\): \(z=p+q\). Set them equal: \(3p=p+q\), so \(q=2p\), or \(p=\dfrac q2\).

Step 3: Check the third corner:
At \((0,3)\): \(z=3q=6p\), which is larger than \(3p\) for \(p>0\). So the minimum is indeed \(3p\), attained at both corners.

Step 4: Why the other options are wrong.
\(p=3q\), \(3p=q\) and \(p=2q\) give different values of \(z\) at the two corners, so the minimum is at one of them only (or at \((0,3)\)).

Final Answer:
p = q/2. \[ \boxed{\text{(C) }p=\dfrac q2} \]