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:

$a=5, b=2$ $a=4, b=5$ $a=2, b=5$ $a=5, b=4$

$a=2, b=5$

$z = ax+ by$ z maximize = 50 at point (0, 10) and (10, 6) $z(0, 10)=0+10b=50$ so $10b=50$ $b=5$ $z(10, 6)=10a+6b=50$ $50=10+6×5$ $50-30=10a$ $10a=20$ $a = 2$