The corner points of the feasible region determined by the following system of linear inequalities: $2x+y \le 10$, $x+3y \le 15$, $x,y \ge 0$ are (0,0), (5,0), (3,4) and (0,5). Let $Z$ = $ax + by$, where $a, b> 0$ Condition on $a$ and $b$ so that the maximum of $Z$ occurs at both (3,4) and (0,5) is
Karnataka PU MQP 2023
Solution
The correct answer is $b=3a$
Explanation
Given $Z$ = $ax + by$ ----(1)
To get condition on $a$ and $b$ so that the maximum of $Z$ occurs at both (3,4) and (0,5), let us put values of $x$ and $y$ in equation (1)
∴ For point (3,4), $Z$ = $a \times 3 + b \times 4$ = $3a + 4b$
For point (0,5), $Z$ = $a \times 0 + b \times 5$ = $ 5b$
Equating both, we get $3a + 4b$ = $ 5b$ ⇒ $3a = b$