Chapter 4, Problem 44RE

To determine

To find : The values of $r$ and $s$ .

Answer to Problem 44RE

The answer:

  $\begin{array}{c}r=25\\ s=50\end{array}$

Explanation of Solution

Given information:

The circle is:

  

Calculation:

Perimeter $=2r+s$

The perimeter is $100\text{ft}$ .

So,

  $2r+s=100\to (1)$

Where $r$ is radius and $s$ is Arc length.

Area $=\pi r^2\times \theta \to (2)$

Where $\theta$ is a angle of arc in radians.

And, $s=2\pi r\times \theta$

Where $\theta$ is a angle of arc in radians.

To use the equation $(1)$:

  $\begin{array}{c}r=50-\frac{s}{2}\\ =50-\pi r\times \theta \end{array}$

Simplify and write $\theta$ as the term of $r$ as:

  $\begin{array}{c}r+\pi r\times \theta =50\\ \Rightarrow r(1+\pi \theta )=50\\ \Rightarrow 1+\pi \theta =\frac{50}{r}\\ \theta =\frac{50-r}{\pi r}\to (3)\end{array}$

To use equation $(3)$ put the value of $\theta$ into equation $(2)$ and get:

  $\begin{array}{c}\text{ Area }=\pi r^2\times \frac{50-r}{\pi r}\\ \text{A}(\text{r})=r\times (50-r)\end{array}$

To find the value of $r$ for that the area is maximum,

To find the value $r$ for which $50r-r^2$ is maximum.

  $A^{\prime }(r)=50-2r$

  $A^{\prime }(r)$ is zero when $r=25$

  $A^{″}(r)=-2$ that is negative.

Therefore, the area is maximum when $r=25$

  $\begin{array}{c}s=2\pi r\times \theta \\ =2\times (50-r)\\ =2(25)\\ =50\end{array}$

Therefore, the required area of the arc to be maximum when $r=25$ and

  $s=50$ .