Solve graphically:
Maximise $Z = 2x + y$ subject to $x + y \leq 1200$, $x + y \geq 600$
$y \leq \frac{x}{2}$, $x \geq 0, y \geq 0.$

2400

The correct answer is Option (3) → 2400 ##

We have, maximize $Z = 2x + y$ subject to

$x + y \leq 1200$

$x + y \geq 600$

$y \leq \frac{x}{2}$

$x \geq 0, y \geq 0$

Let $l_1 : x + y = 1200$

      $l_2 : x + y = 600$

      $l_3 : y = \frac{x}{2} \Rightarrow x = 2y$

The shaded region $ABCD$ is the feasible region and is bounded. The corner points are $A(600, 0), B(1200, 0), C(800, 400), D(400, 200)$.

Hence, $Z$ is maximum at $B(1200, 0)$ i.e., $2400$.