The corner points of the feasible region determined by system of linear constraints are (60, 0), (120, 0), (40, 20) and (60, 30). Let z = ax + by, a, b>0 be the objective function. Find condition on a and b so that the maximum of z occurs at (120, 0) and (60, 30).

The correct option is 2a=b.

We are given the objective function:
$z=ax+by$
where $a,b>0$.

The problem states that the maximum value of $z$ occurs at two corner points: $(120,0)$ and $(60,30)$.

If the maximum value of the objective function occurs at two different corner points, then the value of $z$ at both of these points must be equal.

Let us calculate the value of $z$ at the point $(120,0)$:
$z_1=a\left(120\right)+b\left(0\right)=120a$

Now, let us calculate the value of $z$ at the point $(60,30)$:
$z_2=a\left(60\right)+b\left(30\right)=60a+30b$

Since the maximum value occurs at both points, we set these two values equal to each other:
$120a=60a+30b$

Subtracting $60a$ from both sides of the equation, we get:
$120a-60a=30b$
$60a=30b$

Dividing both sides by 30 to simplify the relation:
$2a=b$

Thus, the required condition on $a$ and $b$ is $2a=b$.