Question

Determine graphically the minimum value of the objective function
$Z=50x+20y...\left(1\right)$
subject to the constraints:
$2xy\ge 5...\left(2\right)$
$3x+y\ge 3...\left(3\right)$
$2x-3y\le 12...\left(4\right)$
$x\ge 0,y\ge 0...\left(5\right)$.

Answer 1

First of all, let us graph the feasible region of the system of inequalities $\left(2\right)$ to $\left(5\right)$. The feasible region (shaded) is shown in the Fig. Observe that the feasible region is unbounded.
We now evaluate $Z$ at the corner points.

Corner Point	$Z=-50x+20y$
$(0,5)$	$100$
$(0,3)$	$60$
$(1,0)$	$-50$
$(6,0)$	$-300\leftarrow smallest$

From this table, we find that $-300$ is the smallest value of $Z$ at the corner point $(6,0)$. Can we say that minimum value of $Z$ is $-300$? Note that if the region would have been bounded, this smallest value of $Z$ is the minimum value of $Z$ (Theorem $2$). 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 graph the inequality
$-50x+20y<-300$ (see Step $3$(ii) of corner Point Method.)
i.e., $-5x+2y<-30$
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 Fig, it has common points.

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