Posted by Salman on Saturday, November 7, 2009 at 3:15pm.
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?

Math  Reiny, Saturday, November 7, 2009 at 4:09pm
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)