math
posted by rolan .
An 8 ft wall stands 27 ft from a building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.

make a sketch.
let x be the distance for the wall to where the beam touches the ground
let the height of the beam along the building be y+8 ft
by similar triangles
y/27 = 8/x
xy = 216 > y = 216/x
let L be the length of the beam
L^2 = (8+y)^2 + (27+x)^2
= (8 + 216/x)^2 + (27+x)^2
2L dL/dx = 2(8+216/x)(216/x^2) + 2(27+x)
= 0 for a min of L
2(8+216/x)(216/x^2) + 2(27+x) = 0
(8+216/x)(216/x^2) + (27+x) = 0
1728/x^2  46656/x^3 + 27+x = 0
times x^3
1728x  46656 + 27x^3 + x^4 = 0
x^3(x+27)  1728(x+27) = 0
(x^3  1728)(x+27) = 0
x = 12 or x = 27, the last one is a "silly" answer
if x=12
L^2 = (8+18)^2 + (27+12)^2
= 2197
L = √2197 = 46.87