Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 66

Suppose that a steel pipe is being carried down a hallway 9ft wide. At the end of the hall, there is a night-angled turn into a narrower hallway 6ft wide. What is the length of the longest pipe that can be carried horizontally around the corner?



The total length of the pipe is $L = L_1 + L_2$

$
\begin{equation}
\begin{aligned}
\sin \theta &= \frac{9}{L_1}\\
\\
L_1 &= \frac{9}{\sin \theta}
\end{aligned}
\end{equation}
$

By using cosine function of $L_2$,

$
\begin{equation}
\begin{aligned}
\cos \theta &= \frac{6}{L_2}\\
\\
L_2 &= \frac{6}{\cos \theta}
\end{aligned}
\end{equation}
$


So, $\displaystyle L = \frac{9}{\sin \theta} + \frac{6}{\cos \theta}$

If we take the derivative with respect to $\theta$, we get...
$\displaystyle \frac{dL}{d\theta} = -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta$
when $\displaystyle \frac{dL}{d\theta} = 0$, then

$
\begin{equation}
\begin{aligned}
0 &= -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta\\
\\
\frac{\sin^3 \theta}{\cos^ \theta} &= \frac{3}{2}\\
\\
\tan^3 \theta &= \frac{3}{2}\\
\\
\tan \theta &= \sqrt[3]{\frac{3}{2}}\\
\\
\theta &= 0.8528
\end{aligned}
\end{equation}
$


Therefore, the length of the pipe is

$
\begin{equation}
\begin{aligned}
L &= \frac{9}{\sin(0.8528)} + \frac{6}{\cos(0.8528)}\\
\\
L &= 21.07\text{ft}
\end{aligned}
\end{equation}
$