calculus
posted by Erica .
A fence 8 feet tall stands on level ground and runs parallel to a tall building. If the fence is 1 foot from the building, find the shortest ladder that will extend from the ground over the fence to the wall of the building.

I assume you made a sketch.
let the distance from the fence to the foot of the ladder be x
let the height of the ladder up the wall be h
by ratios:
8/x = h/(1+x)
hx = 8(1+x
h = (8+8x)/x
let the length of the ladder be L
L^2 = h^2 + (1+x)^2
= (8+8x)^2/x^2 + (1+x)^2
= (64 + 128x + 64x^2)/x^2 + 1 + 2x + x^2
= 64/x^2 + 128/x + 64 + 1 + 2x + x^2
2L dL/dx = 128/x^3  128/x^2 + 2 + 2x
= 0 for a min of L
times x^3
128 + 128x  2x^3  2x^4 = 0
128(1+x)  2x^3(1+x) = 0
(1+x)(2x^3  128) = 0
x = 1 , which makes no sense
or
2x^3  128 = 0
x^3 = 64
x = 4
So L^2 = (8 + 32)^2 /4 + (1+4)^2
= 400 + 25
L^2 = 425
L = √425 = 5√17 = appr 20.6 ft
check my arithmetic. 
This solution is correct but isn't the last step wrong because isn't it (8+32)^2/4^2 instead of (8+32)^4/4?