The corner points of the feasible region determined by the system of Linear Constraints are (0, 3), (1, 2) and (3, 0). Let $z=px+qy, $ where $p, q>0,$ be the objective function. Find the condition on p and q so that the minimum of z occurs at (3, 0) and (1, 2) :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $p=q$
as max. occurs both
on (3, 0) as well as (1, 2)
$Z(1,2)=Z(3,0)$
$p+2q=3p$
$⇒p=q$