Math
posted by Salman .
A fence 4 feet tall runs parallel to a tall building at a distance of 2 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?

Did you make a diagram?
Let the distance from the fence to the foot of the ladder be x feet.
Let the top of the ladder by y feet above the ground.
Let L be the length of the ladder.
L^2 = (x+2)^2 + y^2
by similar triangles:
4/x = y/(x+2)
so L^2 = (x+2)^2 + 16(x+2)^2/x^2
= (x+2)^2(1 + 16/x^2)
2L(dL/dx) = (x+2)^2(32/x^3) + (1+16/x^2)(2)(x+2)
= (x+2)(32(x+2)/x^3 + 2(1+16/x^2)
For a max/min of L, dL/dx = 0
so (x+2)(32(x+2)/x^3 + 2(1+16/x^2) = 0
x = 2 clearly cannot be a solution , so (32(x+2)/x^3 + 2(1+16/x^2) = 0
32/x^2  64/x^3 + 2 + 32/x^2 = 0
2 = 64/x^3
x^3 = 32
x = 32^(1/3)
carefully sub x = 32^(1/3) back into the L^2 equation to find L.
(I got 8.32 feet)
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