## Precalculus (6th Edition) Blitzer

The illumination is $i=2.4\text{ foot-candles}$ at a distance of 50 feet. As per the question $i\propto \frac{1}{{{d}^{2}}}$.
That means $i=\frac{k}{{{d}^{2}}}$ Where i is illumination, d is the distance and k is the constant. Now, calculate the value of k, \begin{align} & i=\frac{k}{{{d}^{2}}} \\ & 3.75=\frac{k}{{{(40)}^{2}}} \\ & k=3.75{{(40)}^{2}} \\ & k=6000 \\ \end{align} Substituting the value of k and d in $i=\frac{k}{{{d}^{2}}}$. \begin{align} & i=\frac{6000}{{{d}^{2}}} \\ & i=\frac{6000}{{{(50)}^{2}}} \\ & \ i=\frac{6000}{2500} \\ & i=2.4\ foot-candles \end{align} Therefore, the illumination is $i=2.4\text{ foot-candles}$ at a distance of 50 feet.