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 \(p, q \gt 0\). The conditions on p and q so that, minimum of Z occurs at (3, 0) and (1, 1) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(p = \displaystyle\frac{q}{2}\)
Values of Z = px + qy at the corner points:
\(Z(0, 3) = 3q\)
\(Z(1, 1) = p + q\)
\(Z(3, 0) = 3p\)
For the minimum to occur at both (3, 0) and (1, 1), Z must have the same value at both corners:
\(3p = p + q\)
\(2p = q\)
\( \displaystyle p = \frac{q}{2} \)
Check that this common value is the minimum: with q = 2p, Z(0, 3) = 6p, which is more than Z(3, 0) = Z(1, 1) = 3p since p > 0.
NCERT Exemplar:
Linear Programming
