6
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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.

pole

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}

$\endgroup$
1
$\begingroup$

what the diagram is saying that since the wire need to hit the ground, the shortest way from one pole to another is equivalent to the second pole being mirrored underground. Then the shortest distance between two points is a straight line.

Look at the drawing that Kastar made. It is accurate.
You have a right angle triangle where one side is the distance between the poles (pull an imaginary line from the point B parallel to the ground so it hits the line formed by the other pole) and the other side is the sum of heights of both poles.

So you have to use Pythagoras to calculate the long side of the triangle

$\endgroup$
1
$\begingroup$

The geometric way nicely described by Kaster is of course best.

We describe a calculus way. Draw an upward going perpendicular $CM$ at the point $C$. Let $\theta=\angle BCM$ and $\phi=\angle ACM$.

Then the rope has length $f(\theta)=15\sec\theta+30\sec\phi$. We also have the relation $15\tan\theta+30\tan\phi=50$.

Find $\frac{df}{d\theta}$. This is $15\sec\theta\tan\theta+30\sec\phi\tan\phi\frac{d\phi}{d\theta}$.

Use $15\tan\theta+30\tan\phi=50$ to obtain $\frac{d\phi}{d\theta}$. Substitute in our expression for $\frac{df}{d\theta}$.

Set $\frac{d\phi}{d\theta}=0$ and simplify. We get after some manipulation $\sin\theta=\sin\phi$.

This just says angle of incidence is equal to the angle of reflection. Photons, tiny though they are, routinely solve the max/min problem.

Remark: We introduced trigonometric functions because they can be useful in more complicated settings. But in your case, let $C$ split the $50$ into two parts, of length $s$ and $t$. Then we want to minimize $f(s)=\sqrt{s^2+225}+\sqrt{t^2+900}$. Differentiate with respect to $s$, using $s+t=50$ to find that $\frac{dt}{ds}=-1$. When we set the derivative equal to $0$, we get $$\frac{s}{\sqrt{s^2+225}}=\frac{t}{\sqrt{t^2+900}},$$ which says precisely that the sine of the angle of incidence is equal to thew sine of the angle of reflection.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.