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:
Answer & explanation
Correct answer: option 3
$z = ax+ by$
z maximize = 50
at point (0, 10) and (10, 6)
$z(0, 10)=0+10b=50$
so $10b=50$
$b=5$
$z(10, 6)=10a+6b=50$
$50=10+6×5$
$50-30=10a$
$10a=20$
$a = 2$