Question

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

$0.75$inch.

(a) Build a model that expresses the ice volume, V, as a

function of the width, x.

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

Short answer

(a) Expression for the volume is$V\left(x\right)=\frac{x^2(8+\mathrm{\pi })}{64}-1.25x$.

(b) the volume of the rink at width $90ft$is$1297.6f{t}^3$

Step-by-step solution

Step 1. Given data

The shape of the community skating rink is a rectangle with semi-circles attached at the ends

length of the rectangle is $20$ less than width x

Ice thickness is $h=0.75$inches

Step 2. Part (a): area of the rectangle

The width of the rectangle is $b=x$

Length of the rectangle $l=2x-20$

Area of rectangle

role="math" localid="1645912582387" $$\begin{gathered} A_r=l\times b \\ {A}_r=(2x-20)\left(x\right) \\ {A}_r=2x^2-20x \end{gathered}$$

Step 3. Area of semicircles

width of the rectangle is the diameter of semicircles

the radius of semicircles is $r=\frac{x}{2}$

Area of a semicircle

$$\begin{gathered} A=\frac{1}{2}{\mathrm{\pi r}}^2 \\ {\mathrm{A}}_{\mathrm{s}1}=\frac{1}{2}\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2 \\ {\mathrm{A}}_{\mathrm{s}1}=\frac{{\mathrm{x}}^2\mathrm{\pi }}{8} \end{gathered}$$

area of both semicircles

${\mathrm{A}}_{\mathrm{s}}=\frac{{\mathrm{x}}^2\mathrm{\pi }}{4}$

Total area

role="math" localid="1645912863922" $$\begin{gathered} A=A_r+A_s \\ A=2x^2-20x+\frac{x^2\mathrm{\pi }}{4} \\ A=\frac{x^2(8+\mathrm{\pi })}{4}-20x \end{gathered}$$

Step 4. volume of the community skating rink

the thickness of ice is

$$\begin{gathered} h=0.75inches \\ h=0.0625 \end{gathered}$$

Volume is a product of area and height

localid="1645913523380" $$\begin{gathered} V=A\times h \\ V\left(x\right)=\left(\frac{x^2(8+\mathrm{\pi })}{4}-20x\right)\left(0.0625\right) \\ V\left(x\right)=\frac{x^2(8+\mathrm{\pi })}{64}-1.25x \end{gathered}$$

Step 5. Part (b)

width is $x=90$

substitute $90$for $x$in volume function

$$\begin{gathered} V\left(x\right)=\frac{x^2(8+\mathrm{\pi })}{64}-1.25x \\ V\left(90\right)=\frac{{90}^2(8+\mathrm{\pi })}{64}-1.25\left(90\right) \\ V\left(90\right)=1410.1-112.5 \\ V\left(90\right)=1297.6 \end{gathered}$$

so volume is $1297.6f{t}^3$.