Chapter 8.5, Problem 42PS

Solution

(a)

To find : The equation related to $x\text{ and }y$ when the inside and outside radii of the walkway is $x\text{ feet and }y\text{ feet}$ respectively and its diagram.

The required equation is $\pi \left(y^2-x^2\right)=600$ .

Given information :

A circular walkway is to be built around a statue in a park. There is enough concrete available for the walkway to have an area of $600$ square feet.

Explanation :

Let the inside and outside radii of the walkway be $x$ feet and $y$ feet respectively.

The following diagram shows the situation:

  

The area of the walkway is the area of the region bounded by outer circle of radius $y$ minus the area of the region bounded by the inner circle of radius $x$ .

  $\begin{array}{c}A=\pi y^2-\pi x^2\\ =\pi \left(y^2-x^2\right)\end{array}$

Since the area for the walkway is $600$ square feet, therefore

  $\pi \left(y^2-x^2\right)=600\mathrm{.......}\left(i\right)$

The required equation is $\pi \left(y^2-x^2\right)=600$ .

(b)

To find : Four possible pairs of dimensions that satisfy the given equation.

The four possible dimensions are $\left(1,\sqrt{1+\frac{600}{\pi }}\right),\left(2,\sqrt{4+\frac{600}{\pi }}\right),\left(3,\sqrt{9+\frac{600}{\pi }}\right)\text{ and }\left(4,\sqrt{16+\frac{600}{\pi }}\right)$ .

Given information :

A circular walkway is to be built around a statue in a park. There is enough concrete available for the walkway to have an area of $600$ square feet. The equation related to $x\text{ and }y$ is $\pi \left(y^2-x^2\right)=600$ .

Explanation :

From the equation $\left(\text{i}\right)$ , we have

  $\begin{array}{c}\left(y^2-x^2\right)=\frac{600}{\pi }\\ y^2=x^2+\frac{600}{\pi }\\ y=\sqrt{x^2+\frac{600}{\pi }}\end{array}$

Substitute $x=1,2,3,4$ . Then

  $\begin{array}{l}{y}_1=\sqrt{1+\frac{600}{\pi }}\\ y_2=\sqrt{4+\frac{600}{\pi }}\\ y_3=\sqrt{9+\frac{600}{\pi }}\\ y_4=\sqrt{16+\frac{600}{\pi }}\end{array}$

Therefore, four possible dimensions are $\left(1,\sqrt{1+\frac{600}{\pi }}\right),\left(2,\sqrt{4+\frac{600}{\pi }}\right),\left(3,\sqrt{9+\frac{600}{\pi }}\right)\text{ and }\left(4,\sqrt{16+\frac{600}{\pi }}\right)$ .

(c)

To graph : The given equation.

Given information :

A circular walkway is to be built around a statue in a park. There is enough concrete available for the walkway to have an area of $600$ square feet.The equation related to $x\text{ and }y$ is $\pi \left(y^2-x^2\right)=600$ .

Graph :

The graph of the equation $\left(i\right)$ is shown below:

  

Interpretation :

Since $x$ is positive, therefore the portion in the first quadrant represents solutions that make sense in this situation.

(d)

To explain : The change in the width of the walkway if both $x\text{ and }y$ increase.

When both $x\text{ and }y$ increases, $\left(y+x\right)$ increases and $\frac{600}{\pi \left(y+x\right)}$ decreases. Therefore , $\left(y-x\right)$ decreases.

Given information :

A circular walkway is to be built around a statue in a park. There is enough concrete available for the walkway to have an area of $600$ square feet.The equation related to $x\text{ and }y$ is $\pi \left(y^2-x^2\right)=600$ .

Explanation :

Given:

  $\begin{array}{c}\pi \left(y^2-x^2\right)=600\\ \pi \left(y+x\right)\left(y-x\right)=600\\ y-x=\frac{600}{\pi \left(y+x\right)}\end{array}$

When both $x\text{ and }y$ increases, $\left(y+x\right)$ increases and $\frac{600}{\pi \left(y+x\right)}$ decreases. Therefore , $\left(y-x\right)$ decreases.