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.
Let L denote the total length of the ladder.
In right angled triangle i.e. ΔAGB we have:
[tex]L^2=h^2+(x+4)^2[/tex]
( Since by using Pythagorean Theorem)
Also, triangle ΔAGB and ΔCDB are similar.
Hence, the ratio of the corresponding sides are equal.
Hence, we have:
[tex]\dfrac{h}{8}=\dfrac{x+4}{x}[/tex]
i.e.
[tex]h=\dfrac{8(x+4)}{x}[/tex]
Hence, on putting the value of h in equation (1) we get:
[tex]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)[/tex]
Now, we need to minimize L.
Hence, we use the method of differentiation.
We differentiate with respect to x as follows:
[tex]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}][/tex]
when the derivative is zero we have:
[tex]2(x+4)[1-\dfrac{256}{x^3}]=0\\\\i.e.\\\\x=-4\ and\ x=\sqrt[3]{256}[/tex]
But x can't be negative.
Hence, we have:
[tex]x=\sqrt[3]{256}[/tex]
Now, on putting this value of x in equation (2) and solving the equation we have:
[tex]L^2=277.14767[/tex]
Hence,
[tex]L=16.6477\ ft.[/tex]
which on rounding to two decimal places is:
[tex]L=16.65\ ft.[/tex]
