Corner points of a feasible bounded region are (0, 10), (4, 2), (3, 7) and (10, 6). Maximum value 50 of objective function z = ax + by occurs at two points (0, 10) and (10, 6). The value of a and b are:

The correct option is a = 2, b = 5.

We are given that the objective function is:
$z=ax+by$
The maximum value of the objective function $z$ is given as 50, and it occurs at two corner points: (0, 10) and (10, 6).

First, let's substitute the point (0, 10) into the objective function equation:
$z=a\left(0\right)+b\left(10\right)=50$
$10b=50$
$b=\frac{50}{10}=5$

Now, let's substitute the second point (10, 6) into the objective function equation with the maximum value of 50:
$z=a\left(10\right)+b\left(6\right)=50$
$10a+6b=50$

Substitute the value of $b=5$ we found earlier into this equation:
$10a+6\left(5\right)=50$
$10a+30=50$
$10a=50-30$
$10a=20$
$a=\frac{20}{10}=2$

Thus, the values are $a=2$ and $b=5$.