Algebra

Linear Programming

Medium

core

The corner points of the bounded feasible region determined by the system of linear inequalities are $(0, 0)$ , $(2, 4)$ , $(0, 5)$ and $(4, 0)$ . If the maximum value of $z = ax + by$ , where $a, b > 0$ occurs at both $(2, 4)$ and $(4, 0)$ , then

$3a = b$

$a = b$

$a = 2b$

$2a = b$

✅ Correct Option: 3

The corner points of the feasible region are $(0, 0)$ , $(2, 4)$ , $(0, 5)$ , and $(4, 0)$ .

The objective function is $z = ax + by$ where $a, b > 0$ .

The maximum value occurs at both $(2, 4)$ and $(4, 0)$ .

In Linear Programming, the maximum value of $z = ax + by$ always occurs at a corner point of the feasible region.

When the maximum occurs at two corner points simultaneously, both points give the same maximum value of $z$ , and the line $z = ax + by$ is parallel to the edge connecting these two points.

Since the maximum occurs at both $(2, 4)$ and $(4, 0)$ , the value of $z$ at these points must be equal.

At point $(2, 4)$ :

$z = a(2) + b(4)$

$z = 2a + 4b$

At point $(4, 0)$ :

$z = a(4) + b(0)$

$z = 4a$

Setting the values equal:

$2a + 4b = 4a$

$4b = 4a - 2a$

$4b = 2a$

$2b = a$

Therefore $a = 2b$ .

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