<s>[INST] Minimizing the length of wire between two poles? There are two poles (lets say poles A and B) $50$ feet apart and the poles are $15$ and $30$ feet tall. There is a wire which runs from the top of pole A to the ground, and then to the top of pole B so that two triangles are made. What is the minimum length of wire needed to set up this configuration? 
I tried finding this by setting up an equation for the length of the wire and taking the first derivative, but I ended up with an ugly polynomial. 
I also tried drawing in one more triangle and used the law of cosines, but that didn't turn out too well either. 
The most likely source of this problem (altough it had different numbers) is from "Maxima and Minima Witouth Calculus" by Ivan Niven, but I'm not sure.
 [/INST] Geometric method
If one mirror reflects $A$ to the $A_1$, then obviously $AC+CB = A_1C+BC$. Latter is minimal when $A_1$, $B$ and $C$ are aligned. One can even find mirror image of $B$ as well, so final $l = \sqrt{(a+b)^2 + h^2}$. One can use another triangles, triangle similarities, etc, but this one is quite visual.

Algebraic method
\begin{align}
l &= \sqrt{a^2+x^2} + \sqrt{b^2+(h-x)^2} \\
\frac {dl}{dx} &= \frac x{\sqrt{a^2+x^2}} - \frac {h-x}{\sqrt{b^2+(h-x)^2}} = 0
\end{align}
from latter one can find that
$$
x^2 \left [ b^2+(h-x)^2\right ] = (a^2+x^2)(h-x)^2 \\
x^2 b^2 + x^2 (h-x)^2 = a^2(h-x)^2 + x^2(h-x)^2 \\
xb = a(h-x)
$$
So, 
\begin{align}
A_0C &= x = \frac {ah}{a+b} \\
B_0C &= h-x = \frac {bh}{a+b}
\end{align}
and
\begin{align}
l &= \sqrt{a^2+\frac {a^2h^2}{(a+b)^2}} + \sqrt{b^2+\frac {b^2h^2}{(a+b)^2}} = \sqrt{1+\frac {h^2}{(a+b)^2}} (a+b) = \sqrt{(a+b)^2+h^2}
\end{align}
</s>