Chapter 3.3, Problem 17PPS

To determine

To find: the coordinate of vertices of feasible regions for the given system of inequalities, $x\ge -6,y+x\le -1\text{and }2x+3y\ge -9$ and find maximum and minimum values of the function $f\left(x,y\right)=-10x-12y$ for this feasible regions.

Answer to Problem 17PPS

  $59$ and $0$ .

Explanation of Solution

Given

Given inequalities are.

  $\begin{array}{l}x\ge -6\\ y+x\le -1\\ 2x+3y\ge -9\\ f\left(x,y\right)=-10x-12y\end{array}$ .

Calculation:

Plotting the graph for given inequalities.

Plotting$x\ge -6$ .

Draw a line parallel to y-axis passing through $x=-6$ ,shade the region right of this line as $x\ge -6$ .

On simplifying $y+x\le -1$ .

  $\begin{array}{l}y+x\le -1\\ y\le -1-x\end{array}$

Plotting $y\le -1-x$ .

Draw graph of the line $y=-1-x$ for the given coordinates below.

  $\begin{array}{cc}x& y\\ 0& -1\\ -1& 0\end{array}$

Shade the region below the line as $y\le -1-x$ .

On simplifying $2x+3y\ge -9$ .

  $\begin{array}{l}2x+3y\ge -9\\ 3y\ge -9-2x\\ y\ge -3-\frac{2}{3}x\end{array}$

Plotting $y\ge -3-\frac{2}{3}x$ .

Draw graph of the line $y=-3-\frac{2}{3}x$ for the given coordinates below.

  $\begin{array}{cc}x& y\\ 0& -3\\ -4.5& 0\end{array}$

Shade the region above the lineas $y\ge -3-\frac{2}{3}x$ .

  

Coordinates of the vertices of feasible region are given in table below.

  $\begin{array}{ccc}\left(x,y\right)& -10x-12y& =f\left(x,y\right)\\ \left(-6,5\right)& -10\left(-6\right)-12\left(5\right)& 0\\ \left(6,-7\right)& 8\left(6\right)-3\left(-7\right)& 59\\ \left(-6,1\right)& 8\left(-6\right)-3\left(1\right)& 51\end{array}$

Therefore, Maximum value is $59$ at $\left(6,-7\right)$ and minimum value is $0$ at $\left(-6,5\right)$ .