Turning a pole/ladder horizontally at a corner I've been attempting to solve the following problem:
A lane runs perpendicular to a road 64 ft wide. If it is just possible to carry a pole 125 ft long from the road into the lane, keeping it horizontal, then what is the minimum width of the lane?
At first, I tried to solve this by thinking that the mid-point of the pole must be at the intersection point of the lane and the road, but that doesn't give the answer, and I have no ideas on how to even attempt this. I have been thinking about it for a while, and finally decided to ask for help.
Any hints/ clues would also be appreciated. Thanks.


From the figure attached, we'll relate the maximum possible pole length $L$ to $w$, the width of the lane.
From the figure
$L = w \csc(\theta) + 64 \sec(\theta)$
for a given $w$, we want to find the minimum attainable length $L$ and this critical value will be our maximum possible pole length $L$
$\dfrac{dL}{d\theta} = - w \csc(\theta) \cot(\theta) + 64 \sec(\theta) \tan (\theta) = 0 $
multiply through by $\sin^2(\theta) \cos^2(\theta) $, then
$0 = - w \cos^3( \theta) + 64 \sin^3(\theta) $
Hence, the critical value for $\theta$ is determined by
$\tan(\theta) = \left( \dfrac{w}{ 64} \right)^{\frac{1}{3}}$
Let $u = \dfrac{w}{64} $ then
$\cos(\theta) = \dfrac{1}{ \sqrt{ 1 + u^\frac{2}{3} }}$
$\sin(\theta) = \tan(\theta) \cos(\theta) = \dfrac{u^\frac{1}{3} } {\sqrt{ 1 + u^\frac{2}{3} } }$
therefore, the maximum $L$ is
$L = (1 + u^\frac{2}{3} )^(\frac{1}{2}) \left(  64 u^\frac{2}{3} + 64 \right) = 64 ( 1 + u^\frac{2}{3} )^\frac{3}{2} $
Since we are given that the maximum $L$ is $125$ then
$ 125 = 64 ( 1 + u^\frac{2}{3} )^\frac{3}{2} $
so that
$ \left(\dfrac{ 125}{64} \right )^{2}{3} = 1 + u^\frac{2}{3} $
This reduces to
$ \dfrac{25}{16} = 1 + u^\frac{2}{3} $
From which,
$ u = \left( \dfrac{9}{16} \right)^\frac{3}{2} = \dfrac{ 27 }{64 } $
But $ u = \dfrac{ w }{64 } $
Therefore
$ w = 27 $
