Chapter 3.2, Problem 42PPS

To determine

To find: The coordinates of the vertices of the figure formed by given system of inequalities

  $\begin{array}{l}2y–x\ge –20\hfill \\ y\ge –3x–6\hfill \\ y\le –2x+2\hfill \\ y\le 2x+14\hfill \end{array}$

Answer to Problem 42PPS

The coordinates of the vertices of a parallelogram ABCD is:

  $A\left(–4,6\right),B\left(1.14,–9.43\right),C\left(4.8,–7.6\right)andD\left(–3,8\right)$

Explanation of Solution

Given information :

Consider the following system of inequalities:

  $\begin{array}{l}2y–x\ge –20\hfill \\ y\ge –3x–6\hfill \\ y\le –2x+2\hfill \\ y\le 2x+14\hfill \end{array}$

Calculation :

Let the inequalities are:

  $\begin{array}{l}2y–x\ge –20\mathrm{.......}\left(1\right)\hfill \\ y\ge –3x–6\mathrm{........}\left(2\right)\hfill \\ y\le –2x+2\mathrm{..........}\left(3\right)\hfill \\ y\le 2x+14\mathrm{............}\left(4\right)\hfill \end{array}$

Use maple to plot the graph of the above equation:

  

The above graph shows a parallelogram ABCD by the four equations.

For the first vertex,

  $\begin{array}{l}y\ge –3x–6\hfill \\ y\le 2x+14\hfill \end{array}$

Write equation (4) as follows,

  $–y\ge –2x–14$

Multiply by 4 in equation (2) and simplify it as follows,

  $\begin{array}{l}\begin{array}{l}\cancel{y}\ge -3x-6\hfill \\ \underset{\_}{\cancel{y}\ge -2x-14}\hfill \end{array}\\ \begin{array}{c}0\ge -5x-20\\ 5x\ge -20\\ x\ge \frac{-20}{5}\\ \ge -4\end{array}\end{array}$

Substitute the points $x=-4$ in equation (1) and simplify it as follows,

  $\begin{array}{l}y\ge -3x-6\\ \begin{array}{cc}y& \ge -3(-4)-6\\ & \ge 12-6\\ & \ge 6\end{array}\end{array}$

The vertex A is $\left(–4,6\right)$

For second vertex,

  $\begin{array}{l}2y–x\ge –20\hfill \\ y\ge –3x–6\hfill \end{array}$

Write equation (4) as follows,

  $y+3x\ge –6$

Multiply by 2 in equation (2) and simplify it as follows,

  $\begin{array}{l}\underset{\_}{\begin{array}{r}\hfill 2y-x\ge -20\\ \hfill 2y+6x\ge -12\\ \hfill (-)(-)\ge (+)\end{array}}\\ \begin{array}{c}0-7x\ge -8\\ x\ge \frac{-8}{-7}\\ \ge 1.14\end{array}\end{array}$

Substitute the points x =1.14 in equation (1) and simplify it as follows,

  $\begin{array}{cc}2y-1.14& \ge -20\\ 2y& \ge -20+1.14\\ y& \ge \frac{-18.86}{2}\\ & \ge -9.43\end{array}$

The vertex B is $\left(1.14,–9.43\right)$

For third vertex,

  $\begin{array}{l}2y–x\ge –20\hfill \\ y\le –2x+2\hfill \end{array}$

Write equation (4) as follows,

  $–y–2x\ge –2$

Multiply by 2 in equation (3) and simplify it as follows,

  $\begin{array}{l}\underset{\_}{\begin{array}{r}\hfill 2y-x\ge -20\\ \hfill -2y-4x\ge -4\end{array}}\\ \begin{array}{r}\hfill 0-5x\ge -24\\ \hfill \begin{array}{l}x\ge \frac{-24}{-5}\\ x\ge 4.8\end{array}\end{array}\end{array}$

Substitute the points $x=4.8$ vin equation (1) and simplify it as follows,

  $\begin{array}{l}\begin{array}{l} 2y–x\ge –20\hfill \\ 2y–4.8\ge –20\hfill \end{array}\\ \begin{array}{l}2y\ge –20+4.8\hfill \\ y\ge \frac{-15.2}{2}\hfill \\ y\ge –7.6\hfill \end{array}\end{array}$

The vertex C is $\left(4.8,–7.6\right)$

For fourth vertex,

  $\begin{array}{l}y\le –2x+2\hfill \\ y\le 2x+14\hfill \end{array}$

Simplify it as follows,

  $\begin{array}{l}y\le 2x+2\hfill \\ \underset{\_}{y\le 2x+14}\hfill \\ 2y\le 16\hfill \\ y\le \frac{16}{2}\hfill \\ y\le 8\hfill \end{array}$

Substitute the points $x=8$ in equation (3) and simplify it as follows,

  $\begin{array}{l}y\le –2x+2\hfill \\ 8\le –2x+2\hfill \\ 8–2\le –2x\hfill \\ \frac{6}{-2}\le x\hfill \\ –3\le x\hfill \end{array}$

The vertex D is $\left(–3,8\right)$

Hence, the coordinates of the vertices of a parallelogram ABCD is:

  $A\left(–4,6\right),B\left(1.14,–9.43\right),C\left(4.8,–7.6\right)andD\left(–3,8\right)$