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A cylindrical can is made from tin.If it can be contain liquid inside it then what is the parameter of design if we are oblige use the minimum amount of tin.

  • calculas,math -

    assuming a constant volume v, we have

    v = pi r^2 h
    so, h = v/(pi r^2)

    the surface area is

    a = 2pi r(r+h)
    = 2pi r(r + v/(pi r^2))
    = 2 pi r^2 + 2v/r

    da/dr = 4pi r - 2v/r^2
    = 2(2pi r^3 - v)/r^2

    we want da/dr=0 for max/min area, so

    r = ∛(v/(2pi))
    h = v/(pi r^2) = ∛(4v/pi)

    but, is this min or max area? Check a'' to be sure it's a minimum

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