Question

Solve the following system of inequation graphically.

$x+2y\le 8,x+y\ge 4,x-y\ge 0,y\ge 0$

Name the common region and write down its vertices

Answer 1

Converting the inequations into equations, we obtain

$x+2y=8,x+y=4,x-y=0,x=0$ and $y=0$

Region represented by $x+2y\le 8$

Its equation form is

$x+2y=8$ $\begin{array}{ccc}x& 8& 0\\ y& 0& 4\end{array}$

The line x+2y=8 meets the coordinates axes at A(8,0) and B(0,4),

respectively. Join these point by a thick line. Clearly, (O,0) satisfies the inequality $x+2y\le 8$.

So, the region contain the origin.

Region represented by $x+y\ge 4$

Its equation form is x+2y=4

$\begin{array}{ccc}x& 4& 0\\ y& 0& 4\end{array}$

The line x+y=4 meets the coordinates axes at C(4,0) and B(0,4) , respectively. Join these points by a thick line .

Clearly, (0,0) does not satisfy the inequality $x+y\ge 4$. So,the region contain the origin.

Region contain the origin.

Region represented by $x-y\le 0$

Its equation form is x-y=0

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

The line x-y=0, i.e. y=x is the line passes through the origin (0,0), (1,1). Join these points by a thick line. Clearly, (0,1) satisfy the inequality. So the region contain the Y-axis. Also $x\ge 0$ and $y\le 0$ so region lies in 1st quadrant. The graph of the given inequalities are shown in figure.

The intersection points of given inequalities are E(8/3, 8/3)and D(2,2).

The vertices of the common region BDE are B(0,4), D(2,2) and $E(\frac{8}{3},\frac{8}{3}).$