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 \(p, q \gt 0\). Condition 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 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 points:
\(3p = p + q\)
\(2p = q\)
\( \displaystyle p = \frac{q}{2} \)
Then the common value is 3p, while Z(0, 3) = 3q = 6p is larger (p > 0), so this value is indeed the minimum.
NCERT Exemplar:
Linear Programming
