Chapter 11, Problem 83RE

To determine

Graph each system of inequalities. Verify whether the graph is bounded or unbounded.

Answer to Problem 83RE

  $\boxed{unbounded}$

Explanation of Solution

Given information:

Graph each system of inequalities. Tell whether the graph is bounded or unbounded, and label the corner points.

  $\{\begin{array}{l}-2x+y\le 2\\ x+y\ge 2\end{array}$

Calculation:

To graph the given inequality, start by graphing the lines $-2x+y=2$ and $x+y=2$ . To determine the part of the graph to be shaded use a test point $\left(0,0\right)$ .

Substitute $x=0,y=0$ in the first inequality of the given system,

  $\begin{array}{l}-2\left(0\right)+\left(0\right)\le 2\\ 0<2\end{array}$

  $\left(0,0\right)$ satisfies the inequlity $-2x+y\le 2$ . shade the part below the line substitute $x=0,y=0$ for second inequality, we get

  $\begin{array}{l}\left(0\right)+\left(0\right)\ge 2\\ 0>2\\ but\\ 0<2\end{array}$

So that $\left(0,0\right)$ does not satisfy the inequality $x+y=2$ .shade the part above the line .

The graph of the given system.

The graph of the system is said to be bounded if it can be contained within a circle of sufficiently large radius, otherwise unbounded.

The corner point of a system are the points of intersection of the boundary lines. the first graph has one corner point, which is $\left(0,2\right)$ .since there is only one intersection point.

Hence, the graph of the system is unbounded and has one corner point. $\boxed{\begin{array}{l}unbounded\\ \left(0,2\right)cornerpo\mathrm{int}\end{array}}$ .