Chapter 3.3, Problem 19PPS

To determine

To find: the system of inequalities, name the coordinate of vertices of feasible regions and find maximum and minimum values of the function for this feasible region.

Answer to Problem 19PPS

  $1344$ and $-486$ .

Explanation of Solution

Given

Given inequalities and function are.

  $\begin{array}{l}-8\le x\le 16\\ y\ge 2x-10\\ 2x+y\le 80\\ f\left(x,y\right)=12x+15y\end{array}$ .

Calculation:

Plotting the graph for given inequalities.

Plotting $-8\le x\le 16$ .

Draw lines parallel to y axis at $x=-8\text{ and }x=16$ and shade the region between the lines.

Plotting $y\ge 2x-10$ .

Draw graph of the line $y=2x-10$ for the given coordinates below.

  $\begin{array}{cc}x& y\\ 0& -10\\ 5& 0\end{array}$

Shade the region above the line $y\ge 2x-10$ .

Simplifying $2x+y\le 80$ .

  $\begin{array}{l}2x+y\le 80\\ y\le 80-2x\end{array}$

Plotting $y\le 80-2x$ .

Draw graph of the line $y=80-2x$ for the given coordinates below.

  $\begin{array}{cc}x& y\\ 0& 80\\ 40& 0\end{array}$

Shade the region below the line $y\le 80-2x$ .

  

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

  $\begin{array}{ccc}\left(x,y\right)& 12x+15y& =f\left(x,y\right)\\ \left(-8,96\right)& 12\left(-8\right)+15\left(96\right)& 1344\\ \left(16,48\right)& 12\left(16\right)+15\left(48\right)& 528\\ \left(16,22\right)& 12\left(16\right)+15\left(22\right)& 522\\ \left(-8,-26\right)& 12\left(-8\right)+15\left(-26\right)& -486\end{array}$

Maximum value is $1344$ at $\left(-8,96\right)$ and minimum value is $-486$ at $\left(-8,-26\right)$ .

So the answer is $1344$ and $-486$ .