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March 27, 2017

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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 ?

  • calculus - ,

    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 = (30-x)/15
    15x = 300-10x
    25x=300
    x = 12

    Put stake 12 m from the 1st post

  • calculus - ,

    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 = (30-x)^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) = -(x-30)/√(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
    (x-12)(x+60) = 0
    x = 12 or x = -60 , but x can't be negative, so

    x = 12

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