Answer
$\text{(a)}\quad V(x)=\frac{\pi}{64}x^2+\frac{1}{8}x^2-\frac{5}{4}x$
$\text{(b)}\quad 1297.61\ \text{ ft}^3$
Work Step by Step
$\text{(a)}$ Let $x$ be the width of the rectangle in feet. Then the radius of the semicircles is $\frac{x}{2}$ feet.
The length $L$ is $20$ feet less than twice the width $(x)$ so $L=2x-20$.
The ice is $0.75$ inch thick, which is $\frac{0.75}{12}$ feet.
The volume $V$ of the ice in the skating rink is:
$\begin{align*}
V(x)&=\text{(thickness of the ice)} \times \text{(area of the skating rink)}\\
&=\frac{0.75}{12}(\text{area of the rectangle}+\text{area of two semicircles})\\
&=\frac{0.75}{12}(xL+\pi r^2)\\
&=\frac{1}{16}\left(x(2x-20)+\pi\left(\frac{x}{2}\right)^2+\right)\\
&=\frac{1}{16}\left(2x^2-20x+\frac{\pi{x^2}}{4}\right)\\
&=\frac{\pi}{64}x^2+\frac{1}{8}x^2-\frac{5}{4}x\\
\end{align*}$
(b) For $x=90\ ft$, we have
$V(90)= \frac{\pi}{64}(90)^2+\frac{1}{8}(90)^2-\frac{5}{4}(90)\approx1297.61\ ft^3$
