359. Choose a place to build a bridge across the river so that the length of the road between two points located on opposite sides of the river is the shortest.
Solution. Let's make a schematic plan of the terrain near the objects specified in the condition (Fig. 58). Distances $a, b, c$ and $h$ are constant according to the condition of the problem. If the bridge is built in the place indicated in the plan, then the length of the road between points $A$ and $B$

$$
l=A C+h+D B
$$

Choosing the distance $A_{1} C$ as the independent variable $x$, we get

$$
A C=\sqrt{a^{2}+x^{2}}, \quad D B=\sqrt{b^{2}+(c-x)^{2}}
$$

II

$$
l=\sqrt{a^{2}+x^{2}}+h+\sqrt{b^{2}+(x-c)^{2}}
$$

where $x$ varies on the segment $[0 ; c]$, which is obvious.

Now let's find the minimum value of the function $l(x)$ on the segment $[0 ; c]$.

We find the derivative $l^{\prime}$ and the critical points lying inside the segment $[0 ; c]$:

$$
\begin{gathered}
l^{\prime}=\frac{x}{\sqrt{a^{2}+x^{2}}}+\frac{x-c}{\sqrt{b^{2}+(x-c)^{2}}}=\frac{x \sqrt{b^{2}+(x-c)^{2}}+(x-c) \sqrt{a^{2}+x^{2}}}{\sqrt{\left(a^{2}+x^{2}\right)\left(b^{2}+\left(x-c^{2}\right)!\right.}} \\
l^{\prime}=0, \text { when } x \sqrt{b^{2}+(x-c)^{2}}+(x-c) \sqrt{a^{2}+x^{2}}=0
\end{gathered}
$$

Solving this equation, we get

$$
\begin{gathered}
x^{2}\left[b^{2}+(x-c)^{2}\right]=(x-c)^{2}\left(a^{2}+x^{2}\right) ; \quad b^{2} x^{2}:=: a^{2}(x-c)^{2} \\
x_{1}=\frac{a c}{a-b} \text { and } x_{2}=\frac{a c}{a+b}
\end{gathered}
$$

The point $x_{1}$ lies outside the segment $0 \leqslant x \leqslant c$: when $a>b, x_{1}>c$; when $a0$ and $\frac{a}{a+b}<1$, i.e., $x_{2}<c$. The derivative $l^{\prime}$ exists everywhere, so the function $l$ has no other critical points. Inside the segment $[0 ; c]$, the function $l$ has only one critical point $x_{2}$. Investigating this critical point by the sign of the derivative $l^{\prime}$ to the left and right of it, as shown in the table, we are convinced that the point $x_{2}$ is a point of minimum.

According to property 1 of continuous functions, in this unique point of minimum on the segment $[0 ; c]$, the continuous function $l$ has the smallest value among all its values on this segment.

Therefore, to make the length of the road between two points located on opposite sides of the river the shortest, the bridge should be built in the place where the distance $A_{1} C=\frac{a c}{a+b}$.