The corner points of the bounded feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let $z = px + qy$ where $p, q> 0$. Then the condition on $p$ and $q$ so that the maximum value of $z$ occurs at (15, 15) and (0, 20) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $q = 3p$
Given: Corner points of feasible region are (0, 10), (5, 5), (15, 15), (0, 20)
Objective function: $z = px + qy$ where $p, q > 0$
To find condition such that maximum value of $z$ occurs at both (15,15) and (0,20).
So, equate the values of $z$ at both points:
$z = p(15) + q(15) = 15p + 15q$
$z = p(0) + q(20) = 20q$
Equating both:
$15p + 15q = 20q$
$15p = 5q$
$\frac{p}{q} = \frac{1}{3}$