Calculus
posted by Anonymous .
A fence 4 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?

Make a diagram
let the foot of the ladder be x ft from the fence
let the ladder reach y ft above the ground
I see similar triangle so set up a ratio
4/x = y/(x+6)
xy = 4x+24
y = (4x+24)/x
let the length of the ladder be L
L^2 = (x+6)^2 + y^2
= (x+6)^2 + [(4x+24)/x]^2
2L dL/dx = 2(x+6) + 2[(4x+24)/x] (x(4)  (4x+24))/x^2
= 2(x+6)  8(x+6)(24)/x^3
= 0 for a min of L
2(x+6)  192(x+6)/x^3 = 0
times x^3
2x^3(x+6)  192(x+6) = 0
2(x+6)(x^3  96) = 0
x = 6 , which makes no sense
or
x = 96^(1/3) , (which is the cuberoot of 96)
x = 4.57886
sub back into L^2 = 197.3
L = 14.05 m