Math calculus

a cylindrical vat must hold 5m^3, the vat must be wider than it is tall, but no more than 3 m in diameter. What dimensions will use the least amount of material?

asked by Jake
  1. v = πr^2h = 5, so
    h = 5/(πr^2)

    a = 2πr^2 + 2πrh
    = 2πr^2 + 10/r

    da/dr = 4πr - 10/r^2
    da/dr=0 when 4πr^3=10
    r = ∛(5/(2π))
    h = 5/(πr^2) = 5/(π(5/(2π))^(2/3)) = ∛(20/π)

    since 3 > 2r > h, it works.

    posted by Steve

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