Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 36

Suppose that a fence 8ft tall runs parallel to a tall building at a distance of 4ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?




Total length of the ladder $ L = L_1 + L_2$, so...

$
\begin{equation}
\begin{aligned}
\sin \theta &= \frac{8}{L_1} \quad ; \quad L_1 = \frac{8}{\sin \theta}\\
\\
\cos \theta &= \frac{4}{L_2} \quad ; \quad L_2 = \frac{4}{\cos \theta}
\end{aligned}
\end{equation}
$


Then,
$\displaystyle L = \frac{8}{\sin \theta} + \frac{4}{\cos \theta} = 8 \csc \theta + 4\sec \theta$
Taking the derivative of $L$ with repsect to \theta, we obtain...
$ L'(\theta) = -8\csc \theta \cot \theta + 4 \sec \theta \tan \theta$

when $L'(\theta) = 0$,

$
\begin{equation}
\begin{aligned}
8 \csc \theta \cot \theta &= 4 \sec \theta \tan\theta\\
\\
\frac{8}{\sin \theta} \left( \frac{\cos \theta}{\sin \theta} \right) &= \frac{4}{\cos \theta} \left( \frac{\sin \theta}{\cos \theta} \right)\\
\\
\frac{\sin^3 \theta}{\cos^3 \theta} &= 2\\
\\
\tan^3 \theta &= 2\\
\\
\theta &= \tan^{-1} [ \sqrt[3]{2} ] = 0.90
\end{aligned}
\end{equation}
$

$L'(0) < 0$ for $\theta$ $\epsilon$ $(0,0.90)$

So, $L$ has an absolute minimum for $\theta = 0.90$, Therefore, the shortest length of the ladder is...
$\displaystyle L = \frac{8}{\sin(0.90)} + \frac{4}{\cos(0.90)} = 16.65$ft.