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Math

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A fence 4 feet tall runs parallel to a tall building at a distance of 2 feet from the building.

What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

  • Math - ,

    Did you make a diagram?
    Let the distance from the fence to the foot of the ladder be x feet.
    Let the top of the ladder by y feet above the ground.
    Let L be the length of the ladder.

    L^2 = (x+2)^2 + y^2

    by similar triangles:
    4/x = y/(x+2)

    so L^2 = (x+2)^2 + 16(x+2)^2/x^2
    = (x+2)^2(1 + 16/x^2)

    2L(dL/dx) = (x+2)^2(-32/x^3) + (1+16/x^2)(2)(x+2)
    = (x+2)(-32(x+2)/x^3 + 2(1+16/x^2)

    For a max/min of L, dL/dx = 0
    so (x+2)(-32(x+2)/x^3 + 2(1+16/x^2) = 0
    x = -2 clearly cannot be a solution , so (-32(x+2)/x^3 + 2(1+16/x^2) = 0
    -32/x^2 - 64/x^3 + 2 + 32/x^2 = 0
    2 = 64/x^3
    x^3 = 32
    x = 32^(1/3)

    carefully sub x = 32^(1/3) back into the L^2 equation to find L.

    (I got 8.32 feet)

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