Turning a pole/ladder horizontally at a corner

Question

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.

Answer 1

The ladder's part already in the lane is not more than $\frac w{\sin\theta}$ long (can you see why?)

At the same time the part still on the road is no longer than $\frac{64\,\mathrm{ft}}{\sin(\frac\pi 2-\theta)}.$

And the sum of those parts must be not less than $125\,\mathrm{ft}$ long.

Can you proceed from here?

Answer 2

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 $