Q.
Assertion (A) The objective function
$Z=-50 x+20 y$
subject to the constraints
$2 x-y \geq-5$
$3 x+y \geq 3$
$2 x-3 y \leq 12$
$x \geq 0, y \geq 0$
has no minimum value.
Reason (R) If the open half plane determined by $a x+b y< m$, where $m$ is the minimum value of $Z$, has a point in common with feasible region, then $Z$ has no minimum value.
Linear Programming
Solution:
Given that
$Z=-50 x+20 y$ ...(i)
subject to the constraints are
$2 x-y \geq-5$ ...(ii)
$3 x+y \geq 3$ ...(iii)
$2 x-3 y \leq 12$ ...(iv)
$x \geq 0, y \geq 0$...(v)
has no minimum value.
First of all, let us graph the feasible region of the system of inequalities (ii) to (v). The feasible region (shaded) is shown in the figure. Observe that the feasible region is unbounded.
We now evaluate $Z$ at the corner points.
Corner point
$z=-50 x+20 y$
$(0,5)$
100
$(0,3)$
60
$(1,0)$
-50
$(6,0)$
-300$\longleftarrow$ Smallest
From this table, we find that $-300$ is the smallest value of $Z$ at the corner point $(6,0)$. Can we say that minimum walue of $Z$ is $-300$ ? Note that, if the region would have been
bounded, this smallest value of $Z$ is the minimum value of $Z$ but here we see that the feasible region is unbounded. Therefore, $-300$ may or may not be the minimum value of $Z$. To decide this issue, we grap the inequality
$-50 x+20 y <-30 $
i.e., $ -5 x+2 y <-30$
and 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, $Z=-50 x+20 y$ has no minimum value subject to the given constraints.
Note In the above example, can you say whether $z=-50 x+20 y$ has the maximum value 100 at $(0,5) ?$ For this, check whether the graph of $-50 x+20 y>100$ has points in common with the feasible region.
| Corner point | $z=-50 x+20 y$ |
|---|---|
| $(0,5)$ | 100 |
| $(0,3)$ | 60 |
| $(1,0)$ | -50 |
| $(6,0)$ | -300$\longleftarrow$ Smallest |