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 :

$b-3 a=0$

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