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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.

  • math -

    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

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