Chapter 6.6, Problem 46PFA

To determine

To find: Area of the triangular region described by the system.

Answer to Problem 46PFA

13.5 square units

Explanation of Solution

Given information:

  $\begin{array}{l}y\le x+5\\ y\le -3x+3\\ y\ge 0\end{array}$

Graph of given inequalities is

From graph we got vertices of Triangle

They are

  $\left(-0.5,4.5\right)$ , $\left(-5,0\right)$ and $\left(1,0\right)$

Area of triangle when vertices are given is

Area $=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

On substituting values

We get

  $\begin{array}{l}\Rightarrow Area=\frac{1}{2}\left|\left(-0.5\right)\left(0-0\right)+\left(-5\right)\left(0-4.5\right)+\left(1\right)\left(4.5-0\right)\right|\\ \Rightarrow Area=\frac{1}{2}\left|0+\left(-5\right)\left(-4.5\right)+\left(1\right)\left(4.5\right)\right|\\ \Rightarrow Area=\frac{1}{2}\left|22.5+4.5\right|\\ \Rightarrow Area=\frac{1}{2}\left|27\right|\\ \Rightarrow Area=13.5\end{array}$

Therefore, area of the triangle formed by the inequalities is 13.5 square units