calculus
posted by erico .
2 posts one is 10m high and other is 15m higs stand 30m apart they are to be stayed by transmission wire attached to a single stake at the ground level, the wire running to the top of the post , where should stake place to used least amount of wire ?

simplest way:
draw a parallelogram ABCD, where AB = 10 (first post)
CD = 15 (2nd post) and BC = 30
Reflect AB in the line BC to point A1 , so that AB= BA1
joint DA1 intersecting BC at P
length of wire = AP+DP = A1P+DP , which is a straight line, and the shortest distance between two points is a straight line, ahhhhh....
let BP=x
Triangles ABP and A1PB are congruent, thus AP=A1P.
Also angles BPA1 and CPD are opposite, thus equal, making triangles APB and DPC similar
Therefore:
x/10 = (30x)/15
15x = 30010x
25x=300
x = 12
Put stake 12 m from the 1st post 
In the above solution, the actual calculation consists of the last 4 lines, the rest explains why it works
otherwise .... CALCULUS
AP^2 = x^2 + 10^2
AP = √(x^2 + 100)
PD^2 = (30x)^2 + 15^2
PD = √(x^2  60x + 1125)
L = √(x^2 + 100) + √(x^2  60x + 1125)
= (x^2 + 100)^(1/2) + (x^2  60x + 1125)^(1/2)
dL/dx = (1/2)(x^2 + 100)^(1/2) (2x) + (1/2)(x^2  60x + 1125)^(1/2) (5x  60)
= 0 for a min of L
x/√(x^2 + 100) = (x30)/√(x^2  60x + 1125)
square both sides
x^2/(x^2 + 100) = (x^2  60x + 900)/(x^2  60x + 1125)
cross multiply
x^4  60x^3 + 900x^2 + 100x^2  6000x + 90000 = x^4  60x^3 + 1125x^2
125x^2 + 6000x  90000=0
x^2 + 48x  720 = 0
(x12)(x+60) = 0
x = 12 or x = 60 , but x can't be negative, so
x = 12