The corner points of feasible region determined by the following system of linear inequalities:
$2x+y≤10, x+3y≤15, x≥0, y≥0$ are $(0, 0), (5, 0), (3, 4)$ and $(0, 5)$
Let $ z= px +qy ,$ when $p, q > 0$ then the relation between p and q, so that maximum of z occurs at both points $(3, 4)$ and $(0, 5)$ is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $3p=q$
$z= px +qy$
At (3, 4): $z= 3p +4q$
At (0, 5): $z= 0p +5q=5q$
$3p+4q=5q$ (∵ At maximum, objective function (z) must yield same value)
$3p=q$