Calculus

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A closed box is to be rectangular solid with a square base. If the volume is 32in^3, determine the dimensions for which the surface area is minimum.

  • Calculus -

    make a sketch
    let the base be x by x, and the height be y
    so x^2 y = 32
    y = 32/x^2

    SA = 2x^2 + 4xy
    = 2x^2 + 4x(32/x^2)
    = 2x^2 + 128/x

    d(SA)/dx = 4x - 128/x^2 = 0 for a min of SA
    4x = 128/x^2
    4x^3 = 128
    x^3 = 32
    x = 32^(1/3) or appr 3.175
    y = 32/3.175^2

    Well , what do you know, it happens to be a perfect cube.

  • Calculus -

    60.48

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