Problem source

A kingdom consists of a vast plane with a central parabolic hill.
	In a vertical cross-section through the centre of the hill, with the $x$-axis horizontal and the $z$-axis vertical, the surface of the plane and hill is given by 
	\[
	z=\begin{cases}
	\dfrac{1}{2a}(a^{2}-x^{2}) & \mbox{ for }\left|x\right|\leqslant a,\\
	0 & \mbox{ for }\left|x\right|>a.
	\end{cases}
	\]
	The whole surface is formed by rotating this cross-section about the $z$-axis. In the $(x,z)$ plane through the centre of the hill, the king has a summer residence at $(-R,0)$ and a winter residence at $(R,0)$, where $R>a.$ He wishes to connect them by a road, consisting of the following segments: 
\begin{itemize}
\item a path in the $(x,z)$ plane joining $(-R,0)$ to $(-b,(a^{2}-b^{2})/2a),$
	where $0\leqslant b\leqslant a.$
\item a horizontal semicircular path joining the two points $(\pm b,(a^{2}-b^{2})/2a),$
	if $b\neq0;$
\item a path in the $(x,z)$ plane joining $(b,(a^{2}-b^{2})/2a)$ to $(R,0).$
	\end{itemiz}
	The king wants the road to be as short as possible. Advise him on his choice of $b.$

Solution source

The path can be broken down into $5$ sections.

1. The section from $(-R,0)$ to $(-a,0)$ which will have distance $R-a$ and is unchangeable.
2. The distance from $(-a,0)$ to $(-b, \frac{a^2-b^2}{2a})$ whose distance we will calculate shortly.
3. The distance from $(-b, \frac{a^2-b^2}{2a})$ to $(b, \frac{a^2-b^2}{2a})$ which will have distance $\pi b$.
4. The distance from $(b, \frac{a^2-b^2}{2a})$ to $(a,0)$ which will have the same distance as 2.
5. The distance from $(a,0)$ to $(R,0)$ which will have distance $R-a$ and we have no control over.

\begin{align*}
\text{distance 2.} &= \int_b^a \sqrt{1 + \left ( \frac{x}{a}\right)^2 } \d x 
\end{align*}
We want to minimize the total, by varying $b$, so it makes sense to differentiate and set to zero.

\begin{align*}
&&0&= -2\sqrt{1+\frac{b^2}{a^2}} + \pi \\
\Rightarrow && \frac{\pi^2}{4} &= 1 + \frac{b^2}{a^2} \\
\Rightarrow && b &= a \sqrt{\frac{\pi^2}{4}-1}
\end{align*}

Since $\pi \approx 3$ this point is outside our range $0 \leq b \leq a$, and our derivative is always positive. Therefore the distance is always increasing and the king would be better off going around the hill as soon as he arrives at it.