Solve the following problem graphically: Minimise and Maximise $Z = 3x + 9y$, Subject to the constraints: $x + 3y \le 60$, $x + y \ge 10$, $x \le y$, $x \ge 0, y \ge 0$.

Min: $60$, Max: $180$

The correct answer is Option (1) → Min: $60$, Max: $180$ ##

Let

$x + 3y \le 60  \dots(i)$

$x + y \ge 10  \dots(ii)$

$x \le y  \dots(iii)$

$x \ge 0, y \ge 0  \dots(iv)$

For $x+3y=60$

For $x+y=10$

From the graph, we see that the feasible region is $ABCDA$. The coordinates of the corner points are $A(0, 10)$, $B(5, 5)$, $C(15, 15)$, and $D(0, 20)$.

Now we evaluate $Z$ at each corner point:

Hence, the maximum value of $Z$ on the feasible region occurs at the two corner points $C(15, 15)$ and $D(0, 20)$ and it is 180 in each case. The minimum value of $Z$ is 60 at the point $B(5, 5)$ of the feasible region.