Maximise $Z = x + y$ subject to $x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$.

3.9

The correct answer is Option (4) → 3.9

We are given that $Z = x + y$ subject to the constraints

$x + 4y \leq 8 \dots(i)$

$2x + 3y \leq 12 \dots(ii)$

$3x + y \leq 9 \dots(iii)$

$x \geq 0, y \geq 0$

On solving eq. $(i)$ and $(iii)$ we get

$x = \frac{28}{11} \text{ and } y = \frac{15}{11}$

Here, $OABC$ is the feasible region whose corner points are

$O(0, 0), A(3, 0), B\left(\frac{28}{11}, \frac{15}{11}\right), C(0, 2)$

Let us evaluate the value of $Z$

Hence, the maximum value of Z is 3.9 at $B(\frac{28}{11},\frac{15}{11})$