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? (Round your answer to two decimal places.) ft

Answer 1

Answer:

The length of the shortest ladder that will reach from the ground over the fence to the wall of the building is:

                              16.65 ft.

Step-by-step explanation:

Let L denote the total length of the ladder.

In right angled triangle i.e. ΔAGB we have:

$L^2=h^2+(x+4)^2$

( Since by using Pythagorean Theorem)

Also, triangle ΔAGB and ΔCDB are similar.

Hence, the ratio of the corresponding sides are equal.

Hence, we have:

$\dfrac{h}{8}=\dfrac{x+4}{x}$

i.e.

$h=\dfrac{8(x+4)}{x}$

Hence, on putting the value of h in equation (1) we get:

$L^2=(\dfrac{8(x+4)}{x})^2+(x+4)^2\\\\i.e.\\\\L^2=\dfrac{64(x+4)^2}{x^2}+(x+4)^2\\\\i.e.\\\\L^2=(x+4)^2[\dfrac{64}{x^2}+1]----------(2)$

Now, we need to minimize L.

Hence, we use the method of differentiation.

We differentiate with respect to x as follows:

$2L\dfrac{dL}{dx}=2(x+4)[\dfrac{64}{x^2}+1]+(x+4)^2\times \dfrac{-128}{x^3}\\\\i.e.\\\\2L\dfrac{dL}{dx}=2(x+4)[\dfrac{64}{x^2}+1+(x+4)\times \dfrac{-64}{x^3}]\\\\\\i.e.\\\\\\2L\dfrac{dL}{dx}=2(x+4)[\dfrac{64}{x^2}+1-\dfrac{64}{x^2}-\dfrac{256}{x^3}]\\\\\\i.e.\\\\\\2L\dfrac{dL}{dx}=2(x+4)[1-\dfrac{256}{x^3}]$

when the derivative is zero we have:

$2(x+4)[1-\dfrac{256}{x^3}]=0\\\\i.e.\\\\x=-4\ and\ x=\sqrt[3]{256}$

But x can't be negative.

Hence, we have:

$x=\sqrt[3]{256}$

Now, on putting this value of x in equation (2) and solving the equation we have:

$L^2=277.14767$

Hence,

$L=16.6477\ ft.$

which on rounding to two decimal places is:

$L=16.65\ ft.$