# A fence 6 feet tall runs parallel to a tall building at a distance of 6 feet 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?

Let x be the distance from where the ladder hits the ground to the fence, and let h be the height on the wall where the ladder hits the wall, and let l be the length of the ladder.

Then using the pythagorean theorem we get :

`l^2=(x+6)^2+h^2`

Also, the...

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Let x be the distance from where the ladder hits the ground to the fence, and let h be the height on the wall where the ladder hits the wall, and let l be the length of the ladder.

Then using the pythagorean theorem we get :

`l^2=(x+6)^2+h^2`

Also, the triangles formed by the wall and the fence are similar, so:

`h/(6+x)=6/x` so `h=36/x+6`

Then

`l=(x^2+12x+72+(12(36))/x+36^2/x^2)^(1/2)`

The ladder length needs to be minimized, so we find `(dl)/(dx)` and set it equal to zero to find the critical points.

`(dl)/(dx)=1/2(x^2+12x+72+(12(36))/x+36/x^2)^((-1)/2)(2x+12-(12(36))/x^2-(2(36)^2)/x^3)`

`=(2x+12-(12(36))/x^2-(2(36)^2)/x^3)/(2(sqrt(x^2+12x+72+(12(36))/x+36^2/x^2)))`

`(dl)/(dx)=0==> 2x+12-(12(36))/x-(2(36)^2)/x^3=0`

`x^4+6x^3-216x+1296=0`

`x^3(x+6)-216(x+6)=0`

`(x+6)(x-6)(x^2+6x+36)=0`

The two real zeros are 6 and -6. Since the length is positive we choose x=6.

Then x=6, the distance from the foot of the ladder to the wall along the ground is 12, the height is 9, so the length of the ladder is 15.

The shortest length for the ladder is 15 ft.

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