Chapter 3.3, Problem 8PPS

To determine

To Find: the coordinate of vertices of feasible regions for the given system of inequalities, $1\le y\le 4,4y-6x\ge -32\text{ and 2}y\ge -x+4$ and find maximum and minimum values of the function $f\left(x,y\right)=-6x+3y$ for this feasible regions.

Answer to Problem 8PPS

  $36$ and $-36$ .

Explanation of Solution

Given

Given inequalities are.

  $\begin{array}{l}1\le y\le 4\\ 4y-6x\ge -32\\ \text{2}y\ge -x+4\\ f\left(x,y\right)=-6x+3y\end{array}$ .

Calculation:

Plotting the graph for given inequalities.

Plotting $1\le y\le 4$ .

Draw two vertical lines parallel to x-axis passing through $y=1\text{and}y=4$ ,shade the region between the lines as shown in red colour.

On simplifying $4y-6\ge -32$

  $\begin{array}{l}4y-6x\ge -32\\ 4y\ge -32+6x\\ y\ge -8+\frac{3}{2}x\end{array}$

Plotting $y\ge -8+\frac{3}{2}x$ .

Draw graph of the line $y=-8+\frac{3}{2}x$ for the coordinates.

  $\begin{array}{cc}x& y\\ 0& -8\\ \frac{16}{3}& 0\end{array}$

Shade the region above the line as $y\ge -8+\frac{3}{2}x$ .

On simplifying the $\text{2}y\ge -x+4$ .

  $\begin{array}{l}\text{2}y\ge -x+4\\ y\ge -\frac{1}{2}x+2\end{array}$

Plotting $y\ge -\frac{1}{2}x+2$ .

Draw graph of the line $y=-\frac{1}{2}x+2$ for the coordinates.

  $\begin{array}{cc}x& y\\ 0& 2\\ 4& 0\end{array}$

Shade the region above the line as $y\ge -\frac{1}{2}x+2$ .

  

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

  $\begin{array}{ccc}\left(x,y\right)& -6x+3y& =f\left(x,y\right)\\ \left(8,4\right)& -6\left(8\right)+3\left(4\right)& -36\\ \left(6,1\right)& -6\left(6\right)+3\left(1\right)& -33\\ \left(2,1\right)& -6\left(2\right)+3\left(1\right)& -9\\ \left(-4,4\right)& -6\left(-4\right)+3\left(4\right)& 36\end{array}$

Therefore, Maximum value is $36$ at $\left(-4,4\right)$ and minimum value is $-36$ at $\left(8,4\right)$ .