Chapter 11, Problem 85RE

To determine

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

Answer to Problem 85RE

  $\boxed{Bounded}$

  $\left(0,0\right),\left(0,2\right)and\left(3,0\right)$

Explanation of Solution

Given information:

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

  $\{\begin{array}{l}x\ge 0\\ y\ge 0\\ x+y\le 4\\ 2x+3y\le 6\end{array}$

Calculation:

Consider the inequality.

  $\{\begin{array}{l}x\ge 0\\ y\ge 0\\ x+y\le 4\\ 2x+3y\le 6\end{array}$

To graph the given inequality, start by graphing the lines $x=0$ , $y=0$ and $2x+3y=6$ . The first equation represents the $y-axis$ and second equation represents $x-axis$

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,

  $\begin{array}{l}\left(0\right)+\left(0\right)\le 4\\ 0\le 4\end{array}$

Since $0\le 4$ satisfy the inequality at point $\left(0,0\right)$ , shade the part below the line .

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

  $\begin{array}{l}2\left(0\right)+3\left(0\right)\le 6\\ 0\le 6\end{array}$

Since $0\le 6$ satisfy the inequality at point $\left(0,0\right)$ , shade the part below the line.

The intersection of the shaded region represents the graph of the given system of inequalities.

The graph of a system is said to be bounded if it can be contained within circle of sufficiently large radius. If not then it is unbounded.

The graph of a system is bounded since it can be contained within a circle.

The corner point of the graph are the point of intersection of the boundary lines. Here we have three corner points, $\left(0,0\right),\left(0,2\right)and\left(3,0\right)$ .

Hence, the graph is $\boxed{Bounded}$ ,and corner points are $\left(0,0\right),\left(0,2\right)and\left(3,0\right)$ .