Solve the following linear programming problem graphically: Maximise $Z = 4x + y$, Subject to the constraints: $x + y \le 50$, $3x + y \le 90$, $x \ge 0, y \ge 0$.

Max $Z = 120$ at $(30, 0)$

The correct answer is Option (2) → Max $Z = 120$ at $(30, 0)$ ##

Let

$ x + y \le 50  \dots(i)$

$3x + y \le 90  \dots(ii)$

$x \ge 0, y \ge 0  \dots(iii)$

For $x + y = 50$:

For $3x + y = 90$:

From the graph, the feasible region is $OABCO$. The coordinates of the corner points are $O(0, 0), A(30, 0), B(20, 30)$ and $C(0, 50)$.

Hence, maximum value of $Z$ is 120 at the point $A(30, 0)$.