Solve graphically:
Maximise $Z = 2x + y$ subject to $x + y \leq 1200$, $x + y \geq 600$
$y \leq \frac{x}{2}$, $x \geq 0, y \geq 0.$
Answer & explanation
Correct answer: option 3
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 $A(600, 0), B(1200, 0), C(800, 400), D(400, 200)$.
|
Corner Points |
Value of $Z=2x+y$ |
|
$A(600, 0)$ |
$1200$ |
|
$B(1200, 0)$ |
$2400 \leftarrow (\text{Maximum})$ |
|
$C(800, 400)$ |
$2000$ |
|
$D(400, 200)$ |
$1000$ |
Hence, $Z$ is maximum at $B(1200, 0)$ i.e., $2400$.