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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