Determine graphically the minimum value of the objective function $Z = -50x + 20y$,
Subject to the constraints: $\begin{aligned} 2x - y &\ge -5 \\ 3x + y &\ge 3 \\ 2x - 3y &\le 12 \\ x \ge 0, y &\ge 0 \end{aligned}$
Answer & explanation
Correct answer: option 4
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:
|
For $2x−y=−5$ |
||
|
$x$ |
0 |
1 |
|
$y$ |
5 |
7 |
|
For $3x+y=3$ |
||
|
$x$ |
0 |
1 |
|
$y$ |
3 |
0 |
|
For $2x−3y=12$ |
||
|
$x$ |
6 |
9 |
|
$y$ |
0 |
2 |
From the graph, we observe that the feasible region is unbounded.
We now evaluate $Z$ at the corner points:
|
Corner Point |
Value of $Z=−50x+20y$ |
|
$A(1, 0)$ |
$-50$ |
|
$B(0, 3)$ |
$60$ |
|
$C(0, 5)$ |
$100$ |
|
$D(6, 0)$ |
$-300$ (Smallest) |
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.