Answer
a) $I(d)=\frac{2550}{d^2}$
b) $102$ foot-candles
c) $2.83$ foot-candles
d) $d\approx 7.14$ feet (see the graph)
Work Step by Step
a) Let $I(d)$ be the illumination of the light in foot-candles and let $d$ be the distance from the light source in feet. Since the illumination varies inversely with the square of the distance, $d^2$, we may write \begin{equation}
\begin{aligned}
I(d)=\frac{k}{d^2},
\end{aligned}
\end{equation} where $k$ is the constant of proportionality that must be determined. Given that $I(5)= 25.5$ foot candles when $d=10$ feet, we can use this information to find $k$ as follows:
\begin{equation}
\begin{aligned}
25.5&= \frac{k}{10^2}\\
25.5\cdot 100&= k\\
2550&= k.
\end{aligned}
\end{equation} We can write
\begin{equation}
\begin{aligned}
I(d)= \frac{2550}{d^2}.
\end{aligned}
\end{equation} b) Find $I(5)$ when $d= 5$.
\begin{equation}
\begin{aligned}
I(5)&= \frac{2550}{5^2}\\
&=102.
\end{aligned}
\end{equation} The illumination of this light at a distance of $5$ feet from the source is $102$ foot-candles.
c) Find $I(d)$ when $d= 30$.
\begin{equation}
\begin{aligned}
I(30)&= \frac{2550}{30^2}\\
&\approx 2.83.
\end{aligned}
\end{equation} The illumination of this light at a distance of $30$ feet from the source is about $2.83$ foot-candles.
d) Let $D= 50$, graph both $D$ and $I(d)$ in the same window. The solution is $d\approx 7.14$ feet from the source. See the graph.
