Question

Solve the following Linear Programming Problem graphically:
Minimize $Z=3x+5y$
Subject to
$x+3y\ge 3,x+y\ge 2,x\ge 0,y\ge 0$

Answer 1

Minimize $Z=3x+5y$
Subject to
$x+3y\ge 3,x+y\ge 2,x\ge 0,y\ge 0$

$x+3y=3$
$\begin{array}{ccc}x& 0& 3\\ y& 1& 0\end{array}$

$x+y=2$
$\begin{array}{ccc}x& 0& 2\\ y& 2& 0\end{array}$
$\begin{array}{cc}\text{Corner points}& \text{Value of}Z=3x+5y\\ (0,2)& 10\\ (\frac{3}{2},\frac{1}{2})& 7\\ (3,0)& 9\end{array}$

As the region that is feasible is unbounded.
Hence, $7$ may or may not be minimum value
of $Z$.
For this, we graph the inequality :
$3x+5y<7$
$3x+5y=7$
$\begin{array}{ccc}x& 0& \frac{7}{3}\\ y& \frac{7}{5}& 0\end{array}$
Since, in the graph,
there is no common point with the
feasible region.

Hence, $Z=7$ is minimum at $(\frac{3}{2},\frac{1}{2})$.