Q. 56 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

The inequalities of the shaded region is

$\left\{\begin{array}{l}x\ge 0\\ y\ge 0\\ 2x-y\le 8\\ 3x+4y\ge 12\\ x\le 5\\ y\le 8\end{array}\right.$

Step 1. Given information

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

Step 2: About the parallel lines

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

here  $k=8$then the equation is$y=8$

It  include origin the inequality is $y\le 8$

The lines parallel to y-axis is $x=k$

here$k=5$ then the equation is $x=5$

it includes the origin the inequality is$x\le 5$

Step 3. Equations from given points

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

Here m=slope 

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

Here there are two lines the points for equation 1 is $(4,0)$& $\left(0,3\right)$.

Slope:$m=\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

The line equation 1 for given points:

 $$\begin{gathered} y-0=-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.(x-4) \\ 4y=-3x+12 \\ 3x+4y=12 \end{gathered}$$ 

Since the shaded region above the line the inequality is $3x+4y\ge 12$. 

step 4.second line equation

the given points are $(4,0)$&$(5,2)$

slope $$\begin{gathered} m=\raisebox{1ex}{$(2-0)$}\!\left/ \!\raisebox{-1ex}{$(5-4)$}\right. \\ m=-2 \end{gathered}$$

the line equation ism $$\begin{gathered} y-0=2(x-4) \\ y=2x-8 \\ 2x-y=8 \end{gathered}$$

the shaded region is above the line so the inequality is $2x-y\le 8$

step 5.the solution

The inequalities of the shaded region is

$\left\{\begin{array}{l}x\ge 0\\ y\ge 0\\ 2x-y\le 8\\ 3x+4y\ge 12\\ x\le 5\\ y\le 8\end{array}\right.$