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 length of the shortest ladder is \(16.65\;{\rm{ft}}\).\(16.65\;ft\)

Step-by-step solution

Given data

Length of the fence=\(8\;{\rm{ft}}\)

Fence to building distance = \(4\;{\rm{ft}}\)

Solution

Draw the diagram:

\(\)\(\)\(\)

From the above diagram,

Assume length of the ladder is\(c\).

The ladder forms a right angle triangle with building side and the ground.

By Pythagorean Theorem,

\({c^2} = {\left( {x + 4} \right)^2} + {y^2}........\left( 1 \right)\)

Use properties of similar tringles, from the triangles\(AED\)and\(ABC\)

\(\begin{array}{l}\frac{8}{x} = \frac{y}{{x + 4}}\\y = \frac{{8x + 32}}{x}\end{array}\)

Substitute this value in the equation (1) as:

\({c^2} = {\left( {x + 4} \right)^2} + \left( {\frac{{8x + 32}}{x}} \right)\)

Find the length of the shortest ladder

Consider the function,

\(\begin{aligned}{c}L\left( x \right) &= {\left( {x + 4} \right)^2} + {\left( {\frac{{8x + 32}}{x}} \right)^2}\\L'\left( x \right) &= 2\left( {x + 4} \right) + 2\left( {\frac{{8x + 32}}{x}} \right)\left( {0 - \frac{{32}}{{{x^2}}}} \right)\\L'\left( x \right) &= 2\left( {x + 4} \right) + 2\left( {\frac{{8x + 32}}{x}} \right)\left( { - \frac{{32}}{{{x^2}}}} \right)\\L'\left( x \right) &= 0\\2\left( {x + 4} \right) + 2\left( {\frac{{8x + 32}}{x}} \right)\left( { - \frac{{32}}{{{x^2}}}} \right) &= 0\\\left( {\frac{{8\left( {x + 4} \right)}}{x}} \right)\left( {\frac{{32}}{{{x^2}}}} \right) &= \left( {x + 4} \right)\end{aligned}\)

Divide by \(\left( {x + 4} \right)\) to simplify as:

\(\begin{aligned}{c}\frac{8}{x}.\frac{{32}}{{{x^2}}} &= 1\\{x^3} &= 32.8\\x &= \sqrt(3){{256}}\\x \approx 6.35\end{aligned}\)

Now perform the first derivative test to check its minimum.

\(\begin{aligned}{l}L'\left( 4 \right) &= 2\left( {5 + 4} \right) + 2\left( {\frac{{8.5 + 32}}{5}} \right)\left( { - \frac{{32}}{{{5^2}}}} \right) &= - 18.86\\L'\left( 4 \right) &= 2\left( {8 + 4} \right) + 2\left( {\frac{{8.5 + 32}}{5}} \right)\left( { - \frac{{32}}{{{5^2}}}} \right) &= 12\end{aligned}\)

So, \(L\left( x \right)\) has minimum when \(x = \sqrt(3){{256}}\)

But\(L\left( x \right)\)was the square of the length of the ladder

\(\begin{aligned}{l}{c^2} &= L\left( {\sqrt(3){{256}}} \right)\\\;\;\;\; &= {\left( {\sqrt(3){{256}} + 4} \right)^2} + {\left( {\frac{{8 \times \sqrt(3){{256}} + 32}}{{\sqrt(3){{256}}}}} \right)^2}\\{c^2} = 277.15\\c \approx 16.65\end{aligned}\)

Hence, the length of the shortest ladder is \(16.65\;{\rm{ft}}\).