Chapter 12.1, Problem 4E

a.

To determine

The labeled diagram of the triangle?

Explanation of Solution

Given information:

The hypotenuse of the right angle triangle is $\sqrt{18}$ .

As per the given problem,

  $\begin{array}{l}\text{Base}=x\\ \text{Height}=y\\ \text{Hypotenuse}=\sqrt{18}\end{array}$

The diagram of the triangle is given below.

  

b.

To determine

The function representing the area of the function

Answer to Problem 4E

The function representing the area of the function is $A=\frac{1}{2}x\sqrt{18-x^2}$

Explanation of Solution

Given:

The hypotenuse of the right angle triangle is $\sqrt{18}$ .

Concept Used:

The Pythagoras theorem for a right angle triangle is given by,

  $\begin{array}{l}{B}^2+P^2=H^2\\ \text{Here,}B\text{is}\text{the}\text{base}\text{of}\text{the}\text{triangle}\\ P\text{is}\text{the}\text{perpendicular}\text{of}\text{the}\text{triangle}\\ H\text{is}\text{the}\text{hypotenuse}\text{of}\text{the}\text{triangle}\end{array}$

The formula of area of triangle is given by,

  $A=\frac{1}{2}\times B\times P$

Calculation:

From Pythagoras theorem,

  $\begin{array}{l}{x}^2+y^2={\sqrt{18}}^2\\ x^2+y^2=18\\ y^2=18-x^2\\ y=\sqrt{18-x^2}\end{array}$

Substitute the values in the formula of area of triangle.

  $\begin{array}{c}A=\frac{1}{2}x\times y\\ =\frac{1}{2}x\sqrt{18-x^2}\end{array}$

Conclusion:

The function representing the area of the function is $A=\frac{1}{2}x\sqrt{18-x^2}$ .

c.

To determine

The value of ${\lim }_{x\to 3}A\left(x\right)$ by completing and observing the given table.

Answer to Problem 4E

The area is maximum at $\sqrt{18}$ .

Explanation of Solution

Given:

The area of triangle is maximum at $x=3$ .

The given table is,

$x$	$2$	$2.5$	$2.9$	$3$	$3.1$	$3.4$	$4$
$A\left(x\right)$

As per the given problem,

  $A\left(x\right)=\frac{1}{2}x\sqrt{18-x^2}$ .

Substitute the values of $x$ from the table.

For $x=2$ ,

  $\begin{array}{c}A\left(x\right)=\frac{1}{2}\left(2\right)\sqrt{18-2^2}\\ =3.741\end{array}$

For $x=2.5$ ,

  $\begin{array}{c}A\left(x\right)=\frac{1}{2}\left(2.5\right)\sqrt{18-{2.5}^2}\\ =4.28\end{array}$

For $x=2.9$ ,

  $\begin{array}{c}A\left(x\right)=\frac{1}{2}\left(2.9\right)\sqrt{18-{2.9}^2}\\ =4.49\end{array}$

For $x=3$ ,

  $\begin{array}{c}A\left(x\right)=\frac{1}{2}3\sqrt{18-3^2}\\ =4.5\end{array}$

For $x=3.1$ ,

  $\begin{array}{l}A\left(x\right)=\frac{1}{2}3.1\sqrt{18-{3.1}^2}\\ =4.48\end{array}$ .

For $x=3.4$ ,

  $\begin{array}{l}A\left(x\right)=\frac{1}{2}3.4\sqrt{18-{3.4}^2}\\ =4.31\end{array}$ .

For $x=3.9$ ,

  $\begin{array}{l}A\left(x\right)=\frac{1}{2}\times 3.9\sqrt{18-{3.9}^2}\\ =3.25\end{array}$ .

For $x=4$ ,

  $\begin{array}{l}A\left(x\right)=\frac{1}{2}\left(4\right)\sqrt{18-4^2}\\ =2.828\end{array}$

$x$	$2$	$2.5$	$2.9$	$3$	$3.1$	$3.4$	$3.9$	$4$
$A\left(x\right)$	$3.74$	$4.28$	$4.49$	4.5	4.48	$4.31$	$3.25$	$2.828$

Conclusion:

From above table it is observed the function attain is maximum value at $x=3$ ,

b.

To determine

The maximum area of the triangle is at $x=3$ .

Answer to Problem 4E

The statement that the maximum area of triangle is at $x=3$ is verified.

Explanation of Solution

The graph is shown below

  

Conclusion:

It is clearly shown from the above graph that the area is maximum at $x=3$