calculus
posted by robert .
a fence 8 ft tall stands on level ground and runs parallel to a tall building. if the fence is 1 ft from the building, find the length of the shortest ladder that will extend from the ground over the fence to the wall of the building. (hint #1: if L represents the length of the ladder, the quantity L is minimized when the quantity L2 is minimized, so you do not have to concern yourself with thte square rootjust minimize L2 and that will minimize L. Hint#2: use similar triangles)
L/ 
/8
/__
..1

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
8/x = y/(x+1)
xy = 8x+8
y = (8x+8)/x
let the length of the ladder be L
L^2 = (x+1)^2 + y^2
= (x+1)^2 + [(8x+8)/x]^2
= x^2+2x+1 + 64 + 128/x + 64/x^2
2L dL/dx = 2x + 2  128/x^2  128/x^3
= 0 for a min of L
2x + 2  128/x^2  128/x^3 = 0
multiply by x^3
2x^4 + 2x^3  128x  128 = 0
2x^3(x+1)  128(x+1) = 0
(x+1)(2x^3  128) = 0
x = 1 , not likely
or
2x^3=128
x^3=64
x=4
sub into L^2
L^2 = 5^2 + 10^2 = 125
I minimized L^2 
12.727 ft

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
8/x = y/(x+1)
xy = 8x+8
y = (8x+8)/x
let the length of the ladder be L
L^2 = (x+1)^2 + y^2
= (x+1)^2 + [(8x+8)/x]^2
= x^2+2x+1 + 64 + 128/x + 64/x^2
2L dL/dx = 2x + 2  128/x^2  128/x^3
= 0 for a min of L
2x + 2  128/x^2  128/x^3 = 0
multiply by x^3
2x^4 + 2x^3  128x  128 = 0
2x^3(x+1)  128(x+1) = 0
(x+1)(2x^3  128) = 0
x = 1 , not likely
or
2x^3=128
x^3=64
x=4
sub into L^2
L^2 = 5^2 + 10^2 = 125
I minimized L^2