Question

A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

Short answer

The required length of the ladder is: \[16.65{\rm{ ft}}\].

Step-by-step solution

Optimization of a Function

The Optimization of a Function \[f\left( x \right)\] for any real value of \[c\]can be given as:

\[\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\].

Optimizing the function according to given condition.

The given data in question can be assembled as shown in figure below:

where,

\[\begin{aligned}{l}L \to {\rm{length of ladder}}\\\theta \to {\rm{angle between ladder and ground}}\end{aligned}\]

Now, the length of the ladder can be expressed as:

\[L = 8\,{\rm{cosec}}\theta + 4\sec \theta ,\,\,\,\,\,\left\{ {\forall \theta \in \left( {0,\frac{\pi }{2}} \right)} \right\}\]

For minimization, we have:

\[\begin{aligned}{c}L'\left( \theta \right) = 0\\\frac{d}{{d\theta }}\left( {8\,{\rm{cosec}}\theta + 4\sec \theta } \right) = 0\\ - 8\,{\rm{cosec}}\theta \cot \theta + 4\sec \theta \tan \theta = 0\\2\,{\rm{cosec}}\theta \cot \theta = \sec \theta \tan \theta \\{\tan ^3}\theta = 2\\\tan \theta = {2^{\frac{1}{3}}}\\\theta = {\tan ^{ - 1}}{2^{\frac{1}{3}}}\end{aligned}\]

Now, we have:

\[\begin{aligned}{l}{\rm{for }}0 < \theta < {\tan ^{ - 1}}{2^{\frac{1}{3}}} \to L'\left( \theta \right) < 0\\{\rm{for }}{\tan ^{ - 1}}{2^{\frac{1}{3}}} < \theta < \frac{\pi }{2} \to L'\left( \theta \right) > 0\end{aligned}\]

Therefore, the function will have absolute minimumvalue at \[\theta = {\tan ^{ - 1}}{2^{\frac{1}{3}}}\]. For this, we have the minimum length as:

\[\begin{aligned}{c}L\left| {_{\min }} \right. = 8{\rm{ cosec}}\left( {{{\tan }^{ - 1}}{2^{\frac{1}{3}}}} \right) + 4\sec \left( {{{\tan }^{ - 1}}{2^{\frac{1}{3}}}} \right)\\ = 16.65{\rm{ ft}}\end{aligned}\]

Hence, the required length is: \[16.65{\rm{ ft}}\].