Chapter 2, Problem 12CT

A community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length of the rectangle is $20$ feet less than twice the width. The thickness of the ice is $2$ inches.

Build a model that expresses the ice volume $V$ as a function of the width, $x.$

How much ice is in the rink if the width is $90$ feet?

To determine

The model that expresses the ice volume, $V$, as a function of width, $x$, where a community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length is $20$ feet less than twice the width. The thickness of the ice is 2 inches.

Answer to Problem 12CT

Solution:

The model that expresses the ice volume, $V$, is $V=\frac{1}{6}\left((2x-20)x\text{+}\frac{\pi x^2}{4}\right)$.

Explanation of Solution

Given Information:

The community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length is $20$ feet less than twice the width. The thickness of the ice is 2 inches.

Explanation:

Let $x$ be the width of the rink.

The length of the rink is $20$ feet less than twice the width.

So, the length of the width is $2x-20$.

Therefore, area of rectangle $=\text{Length}\cdot \text{width}$

$\Rightarrow \text{Area of rectangle}=(2x-20)x$

The rink is enclosed by semicircles at the ends.

These two semicircles form a circle.

As width is $x$, the radius of circle will be $\frac{x}{2}$.

Therefore, the area of circle is $\pi {(radius)}^2=\pi {\left(\frac{x}{2}\right)}^2=\frac{\pi x^2}{4}$.

Thus, total area of the rink is the sum of the area of the rectangle and the area of the circle.

$\Rightarrow \text{Area of rink}=\text{Area}\text{of}\text{rectangle}\text{+}\text{Area}\text{of}\text{circle}$

$\Rightarrow \text{Area of rink}=(2x-20)x\text{+}\frac{\pi x^2}{4}$

Thickness of ice is 2 inches.

As other dimensions are in feet, convert 2 inches to feet.

$2\text{inches}=\frac{2}{12}\text{feet}\text{=}\frac{1}{6}\text{feet}$.

Ice volume $V=\text{Area}\text{}\text{}\text{}\text{}\text{}\text{of}\text{rink}\cdot \text{Thickness}$.

Ice volume $V=\left((2x-20)x\text{+}\frac{\pi x^2}{4}\right)\cdot \frac{1}{6}$.

Therefore, the model that expresses the ice volume, $V$, is $V=\frac{1}{6}\left((2x-20)x\text{+}\frac{\pi x^2}{4}\right)$ .

To determine

To calculate: How muchice is in the rink if the width is $90$ feet .

Answer to Problem 12CT

Solution:

When width is $90$ feet, ice in rink is $3460.29f{t}^3$.

Explanation of Solution

Given Information:

Acommunity skating rink is in the shape of a rectangle with semicircles attached at the ends. The length is $20$ feet less than twice the width. The thickness of the ice is 2 inches.

Also, width is $90$ feet.

Formula used:

$\text{Volume}=\text{Area}\cdot \text{Thickness}$

Calculation:

From part (a), model representing ice volume, $V$, is $V=\frac{1}{6}\left((2x-20)x\text{+}\frac{\pi x^2}{4}\right)$.

Substitute $x=90$ in above model,

$\Rightarrow V=\frac{1}{6}\left((2\cdot 90-20)\cdot 90\text{+}\frac{\pi {(90)}^2}{4}\right)$

Simplify above equation,

$\Rightarrow V=\frac{1}{6}\left((180-20)\cdot 90\text{+}\frac{8100\pi }{4}\right)$

$\Rightarrow V=\frac{1}{6}\left(160\cdot 90\text{+}\frac{8100\pi }{4}\right)$

$\Rightarrow V=\frac{1}{6}\left(14400\text{+}\frac{8100\pi }{4}\right)$

$\Rightarrow V=\frac{1}{6}\left(14400\text{+}\frac{8100\pi }{4}\right)$

$\Rightarrow V=\frac{1}{6}\left(20761.725\right)$

$\Rightarrow V=3460.2875f{t}^3$

$\Rightarrow V\approx 3460.29f{t}^3$

Therefore, $3460.29f{t}^3$ ice is in the rink when width is $90$ feet.