Chapter 1.9, Problem 69E

Solution

a)

To find: The area of the region determined by pair of inequalities $y\le -x^2+4x,y\ge 0$ .

Area of the region determined by pair of inequalities $y\le -x^2+4x,y\ge 0$ is equal to $10.7$ square units.

Given:

Figure is as follows.

  

Calculation:

In order to find the area, first graph $y\le -x^2+4x$ . For this, graph $y=-x^2+4x$ .

Substitute some value for $x$ , say, $0$ and find the corresponding value for $y$ .

  $\begin{array}{c}y=-{\left(0\right)}^2+4\left(0\right)\\ =0\end{array}$

Organize the results in a table.

  $\begin{array}{cccccc}x& 0& 1& 2& 3& 4\\ y& 0& 3& 4& 3& 0\end{array}$

Plot these points and join them using a smooth curve. Since the inequality is $\le$ use a solid line to draw the curve.

  

Test a point inside the parabola, say, $\left(2,-4\right)$ .

  $\begin{array}{c}y\le -x^2+4x\\ $ -4\le -{(2)}^2+4(2)$-4\le 4\end{array}$

Thus, $\left(2,-4\right)$ is a solution of the inequality.

Shade the region inside the parabola since the test point is a solution of the inequality.

  

Now, graph $y\ge 0$ . For this, first, graph $y=0$ . Since the inequality is $\ge$ , use a solid line to draw the graph.

  

Test a point inside the graph, say, $\left(2,2\right)$ .

  $\begin{array}{l}y\ge 0\\ 2\ge 0\end{array}$

Thus, $\left(2,2\right)$ is a solution of the inequality.

Shade the region containing the test point as the test point is a solution of the inequality.

  

Identify the purple region where the two graphs intersect. This region is the graph of the system.

  

Compare the above graph with the given diagram, to get $b$ and $h$ as $4$ .

  

Find the area by substituting the values of $b$ and $h$ in $A=\frac{2}{3}bh$ .

  $\begin{array}{c}A=\frac{2}{3}\left(4\right)\left(4\right)\\ \approx 10.7\end{array}$

Thus, the area of the region determined by the given pair of inequality is about $10.7$ square units.

b)

To find: The area of the region determined by pair of inequalities $y\ge x^2-4x-5\text{,}y\le 3$ .

The area of the region determined by pair of inequalities $y\ge x^2-4x-5\text{,}y\le 3$ is equal to $42$ square units.

Given:

Figure is as follows.

  

Calculation:

In order to find the area, repeat the process in part $\left(a\right)$ and identify the region where the two graphs intersect. This region is the graph of the system.

  

Compare the above graph with the given diagram.

Find the area by substituting the values of $b$ and $h$ in $A=\frac{2}{3}bh$ .

  $\begin{array}{c}A=\frac{2}{3}\left(7\right)\left(9\right)\\ \approx 42\end{array}$

Therefore, the area of the region determined by the given pair of inequality is about $42$ square units.