Question

Minimize: $Z=6x+4y$
Subject to : $3x+2y\ge 12$,
$x+y\ge 5$,
$0\le x\le 5$,
$0\le y\le 6$

Answer 1

The feasible region is $DEAY$ which is shaded in the figure.
The vertices of the feasible region are $A(0,6)$ and $D(5,0)$.
$E$ is the point of intersection of the lines. Now solving equation $\left(i\right)$ and $\left(ii\right)$ we get intersecting point $E(2,3)$. Origin do not satisfied given condition $(0,0)$ does not present in under lined area.
Calculation of minimum value:

Corner points	$Z=6x+4y$
At $A(0,6)$	$Z=6\times 0+4\times 6=24$
At $E(2,3)$	$Z=6\times 2+4\times 3=24$
At $D(5,0)$	$Z=6\times 5+4\times 0=30$

Since $Z$ has minimum value $24$ at two consecutive vertices $A$ and $E$ of the feasible region, $Z$ is minimum value $24$ at every point of segment $AE$. Hence, there are infinite number of optimal solutions.