# What is the length of the ladder if a ladder of length L is carried horizontally around a corner from a hall 3 feet wide into a hall 4 feet wide?

Mar 1, 2015

Consider a line segment running from $\left(x , 0\right)$ to $\left(0 , y\right)$ through the interior corner at $\left(4 , 3\right)$.
The minimum length of this line segment will be the maximum length of ladder that can be maneuvered around this corner. Suppose that $x$ is beyond $\left(4 , 0\right)$ by some scaling factor, $s$, of 4, so
$x = 4 + 4 s = 4 \left(1 + s\right)$
[watch for the $\left(1 + s\right)$ showing up later as a value to be factored out of something.]

By similar triangles we can see that
$y = 3 \left(1 + \frac{1}{s}\right)$

By the Pythagorean Theorem, we can express the square of the length of the line segment as a function of $s$
${L}^{2} \left(s\right) = {3}^{2} \left({s}^{- 2} + 2 {s}^{- 1} + 1\right) + {4}^{2} \left(1 + 2 s + {s}^{2}\right)$

Normally we would take the derivative of L(s) to find the minimum but in this case it is easier to take the derivative of ${L}^{2} \left(s\right)$.
(Note that if $L \left(s\right)$ is a minimum as $s = {s}_{0}$, then ${L}^{2} \left(s\right)$ will also be a minimum at $s = {s}_{0}$.)

Taking the first derivative of ${L}^{2} \left(s\right)$ and setting it to zero, we get:
${3}^{2} \left(- 2 {s}^{- 3} - 2 {s}^{- 2}\right) + {4}^{2} \left(2 - 2 s\right) = 0$

Multiplying by ${s}^{3}$ and then factoring out $2 \left(1 + s\right)$
allows us to solve for $s$

$s = {\left(\frac{3}{4}\right)}^{\frac{2}{3}}$

Plugging this value back into the equation for ${L}^{2} \left(s\right)$ and taking the square root (I used a spreadsheet), we get
the maximum ladder length $= 9.87 f e e t$ (approx.)