Calculate minimum width of lane 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 should be the minimum width of the lane?
I am unable to understand what properties should hold for the pole to make its way into the lane. A well-explained solution is preferred (I am a twelfth standard student).
 A: This is a version of the moving ladder problem, though more difficult because the road and lane have different widths. Let $\alpha$ be the angle between the pole and the direction of the road. In the process of moving, $\alpha$ takes on values between $0$ and $\pi/2$.  
Only $64/\sin \alpha$ of the length of the pole can fit within the road. And if the width of the lane is $x$, then $x/\cos \alpha$ of the pole will fit within the lane.  Therefore, we must have 
$$\frac{64}{\sin\alpha} + \frac{x}{\cos\alpha}\ge 125\tag1$$
for all $\alpha$ between $0$ and $\pi/2$. Rearrange as
$$x\ge 125\cos\alpha -  64\cot \alpha \tag2$$
The optimal value of $x$ (when the pole just fits) is the  maximum of the function on the right of (2) on the interval $(0, \pi/2)$. 
A: Let $L$ be the width of the lane. Consider the segment of straight line passing through one corner of the lane and the road, going from the opposite side of the road to the opposite side of the lane. Suppose segment makes an angle $\theta\in (0,\frac{\pi}{2})$ with the road. The lenght of such segment is a function $f$ of $\theta$: $$f(\theta)=\frac{L}{\cos \theta}+\frac{64}{\sin \theta}$$ where $L$ is the width of the lane. Note that $$\lim_{\theta \to 0}f(\theta)=\lim_{\theta \to \frac{\pi}{2}}f(\theta)=+\infty$$ So $f$ has a minimum in $(0,\frac{\pi}{2})$
Let $L$ be the minimun width of the lane such that a pole 125 feet long from the road into the lane, keeping it horizontal. Then, when we minimize $f$ we will find an angle $\alpha\in (0,\frac{\pi}{2})$, such that $$\frac{L}{\cos \alpha}+\frac{64}{\sin \alpha}=125 \qquad \qquad(1)$$ On the other hand, we will also have $f'(\alpha)=0$, which means $$\frac{\sin \alpha}{\cos^2 \alpha}L-\frac{\cos \alpha}{\sin^2 \alpha}64=0$$ So $$L=\frac{\cos^3 \alpha}{\sin^3 \alpha}64$$ Replacing $L$ in $(1)$ and multiplying by $\sin^3 \alpha$, we have
$$64 \cos^2 \alpha + 64 \sin^2 \alpha =125 \sin^3 \alpha$$ So $$64 =125 \sin^3 \alpha$$ So $$ \sin \alpha =\frac{4}{5} \qquad \textrm{ and }\qquad \cos \alpha =\frac{3}{5}$$ Replacing them in $(1)$, and dividing by $5$, we get
$$\frac{L}{3}+\frac{64}{4}=25$$ So $$L=27$$ The minimum width of the the lane is $27$ feet.
