Determine graphically the minimum value of the objective function $Z = -50x + 20y$,

Subject to the constraints: $\begin{aligned} 2x - y &\ge -5 \\ 3x + y &\ge 3 \\ 2x - 3y &\le 12 \\ x \ge 0, y &\ge 0 \end{aligned}$

No minimum value

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

Let

$\begin{aligned} 2x - y &\ge -5 & \dots(i) \\ 3x + y &\ge 3 & \dots(ii) \\ 2x - 3y &\le 12 & \dots(iii) \\ x \ge 0, y &\ge 0 & \dots(iv) \end{aligned}$

Boundary lines tables:

From the graph, we observe that the feasible region is unbounded.

We now evaluate $Z$ at the corner points:

From the table, the smallest value of $Z$ is $-300$, but because the feasible region is unbounded, $-300$ may or may not be the minimum value.

To decide this, we graph the inequality:

$-50x + 20y < -300$

$\text{i.e.,} \, -5x + 2y < -30$

Now, check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then -300 will not be the minimum value of Z. Otherwise, -300 will be the minimum value of Z.

As shown in the figure, it has common points.

Therefore, $Z = -50x + 20y$ has no minimum value subject to the given constraints.