Solve the following Linear Programming Problem graphically: Maximise: $Z = -x + 2y$, Subject to the constraints: $x \ge 3, x + y \ge 5, x + 2y \ge 6, y \ge 0$.

No maximum value

The correct answer is Option (4) → No maximum value ##

The feasible region determined by the constraints, $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$ is given below.

Here, it can be seen that the feasible region is unbounded.

The values of $Z$ at corner points $A(3, 2)$, $B(4, 1)$ and $C(6, 0)$ are given below.

Since the feasible region is unbounded, $Z = 1$ may or may not be the maximum value.

Now, we draw the graph of the inequality, $-x + 2y > 1$, and we check whether the resulting open half-plane has any point/s in common with the feasible region or not.

Here, the resulting open half-plane has points in common with the feasible region.

Hence, $Z = 1$ is not the maximum value. We conclude $Z$ has no maximum value.