Question

In Problems 53–56, write a system of linear inequalities for the given graph.

Short answer

The inequalities of the shaded region is

$\left\{\begin{array}{l}x\ge 0\\ y\ge 0\\ x+y\le 50\\ x-y\le 0\\ x\le 20\\ y\ge 15\end{array}\right.$

Step-by-step solution

Step 1. Given information

The graph is in first quadrant so the values are$x\ge 0$&$y\ge 0$.

About the parallel lines

The lines parallel to x-axis is$y=k$

here $k=15$then the equation is$y=15$

It does not include origin $y\ge 15$

The lines parallel to y-axis is $x=k$

here $k=20$then the equation is $x=20$

it includes the origin$x\le 20$

Step 3. Equations from given points

line equation $y-y_1=m(x-x_1)$

here m=slope

$m=(\raisebox{1ex}{$y_2-y_1$}\!\left/ \!\raisebox{-1ex}{$(x_2-x1)$}\right.$

There are two lines the given points for equation 1 is $(15,15)$&$(20,20)$

slope $$\begin{gathered} m=\raisebox{1ex}{$(15-20)$}\!\left/ \!\raisebox{-1ex}{$(15-20)$}\right. \\ m=\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$-5$}\right. \\ m=1 \end{gathered}$$

line equation islocalid="1646940764105" $$\begin{gathered} y-15=1(x-15) \\ y-15=x-15 \\ x-y=0 \end{gathered}$$

since the shaded region is above the x-axis$x-y\le 0$

Step 4. Second-line equation

The given points are &$(20,30)$

Slope $$\begin{gathered} m=\raisebox{1ex}{$(30-50)$}\!\left/ \!\raisebox{-1ex}{$(20-0)$}\right. \\ m=\raisebox{1ex}{$-20$}\!\left/ \!\raisebox{-1ex}{$20$}\right. \\ m=-1 \end{gathered}$$

The line equation is $$\begin{gathered} y-50=-1(x-0) \\ y-50=-x \\ x+y=50 \end{gathered}$$

The origin includes in the lines the inequalities are $x+y\le 50$.

Step 5. Solution

The inequalities of the shaded region

$\left\{\begin{array}{l}x\ge 0\\ y\ge 0\\ x+y\le 50\\ x-y\le 0\\ x\le 20\\ y\ge 15\end{array}\right.$