Afence 6 feet tall runs parallel to a tall building at a distance of 5 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?

Generally a 45 degree angle is the best angle.  this yields the smallest triangle with a 6' fence vertical side.  As the ladder reaches the wall, two similar triangles are resolved.  the 45 degree angle that the ladder makes with the ground yields a 6x6 legged right triangle.  Adding 5' to the ground makes the larger 45 degree triangle with a base of 5+6=11' side.  Using the formula for Special Triangles, 45-45-90 gives the ladder a 11 times the Squareroot of 2 for its length.

Looks too ugly to me, but this is what I get:

First, draw the triangle of the ladder, with the fence and the building.

There are two similar right triangles: the one from the ground to the fence, and the one from the ground to the building. Let's call L the length of the ladder and l, the length of the ladder from the ground to the fence, then:

L/(x+d) = l/x, where x is the horizontal distance from the point where the ladder touches the ground to the base of the fence and 'd' is the distance between the fence and the building (d=5 ft).

L = (x+d)/x*l, now l = sqrt( x^2+h^2), where h is the height of the fence (h=6 ft).

Then, one usually squares it, because the point where the maximum or minimum is attained is the same when optimizing a distance.

So that, one could look for the minimum of L^2 = (x+d)^2/x^2*(x^2+h^2), where d = 5 ft, and h = 6ft.

What I don;t like, is that L' = dL/dx is a bit lengthy to calculate and then you have to set it = 0. I think all this is right, but I expected a somewhat simpler expression, unless you're doing AP or similar.
 

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