Chapter 3.7, Problem 32WE

To determine

To draw: The graph of given inequality in a coordinate plane.

Answer to Problem 32WE

  

Explanation of Solution

Given information: The given inequality is:

  $\begin{array}{l}y\ge 0(i)\\ y\le x(ii)\\ x+y\le 3(iii)\end{array}$

Calculation: The given inequality is:

  $\begin{array}{l}y\ge 0(i)\\ y\le x(ii)\\ x+y\le 3(iii)\end{array}$ .

From equation $y\ge 0$ :

The solution region is the area above and including the solid boundary line $y=0$ .

The corresponding equation is:

  $\begin{array}{l}y=x(iv)\\ x+y=3(v)\end{array}$

  $\begin{array}{l}y=x\\ \mathrm{put}x=0\\ y=0\\ \mathrm{put}x=1\\ y=1\end{array}$

$x$	$0$	$1$
$y$	$0$	$1$

  $\begin{array}{l}x+y=3\\ \mathrm{put}x=0\\ 0+y=3\\ y=3\\ \mathrm{put}y=0\\ x+0)=3\\ x=3\end{array}$

$x$	$0$	$3$
$y$	$3$	$0$

  

Put $x=0\mathrm{and}y=0$ in equation $(i)$ we get:

  $0\ge 0$

Which is true, therefore the closed plane containing the origin is graph of given inequality.

Put $x=0\mathrm{and}y=0$ in equation $(ii)$ we get:

  $0\le 0$

Which is true, therefore the closed plane containing the origin is graph of given inequality.

Put $x=0\mathrm{and}y=0$ in equation $(iii)$ we get:

  $\begin{array}{l}0+0\le 3\\ 0\le 3\end{array}$

Which is true, therefore the closed plane containing the origin is graph of given inequality.

Thus the common shaded region is the graph of all inequality.