For an objective function $z=a x+b y$ where $a, b>0$. The corner points of feasible region determined by a set of constraints are $(0,20)(10,10)(30,30)$ and $(0,40)$. The condition on a and b such that $\max z$ occurs at both the points $(30,30)$ and $(0,40)$ is :
Answer & explanation
Correct answer: option 1
z = ax+ by a, b > 0
Corner points
(0, 20)
(10, 10)
(30, 30)
(0, 40)
for z max to occur on both (30, 30) and (0, 40)
z(x, y) = ax + by
So z(30, 30) = z(0, 40)
⇒ 30a + 30b = 40b
⇒ 3a+ 3b = 4b (dividing eq by 10)
So 4b - 3b - 3a = 0
b - 3a = 0