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cipker
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cipker 發問於 科學及數學數學 · 1 十年前

AM>=GM

Show that the surface area S, of a closed cylinder of volume V, with base radius r is:

S = 2(pi)r^2 +2V/r

Using AM>= GM only,

Show that:

S/[6(pi)] >= [V^2 / 4(pi)^2]^(1/3)

If the volume is fixed, what is the ratio of the base radius r to the height h when the surface area is a minimum?

1 個解答

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  • 1 十年前
    最愛解答

    S = 2πr^2 + 2πrh

    S = 2πr^2 + 2 (πr^2)h / r

    S = 2πr^2 + 2 V / r

    ;;;;;;

    S / (6π) >= ∛ ( V^2 / 4π^2 )

    [2πr^2 + 2 (πr^2)h / r] / (6π) >= ∛ [ (πr^2 h)^2 / 4π^2) ]

    ( r^2 + rh ) / 3 >= ∛ [ (r^4) (h^2) / 4 ]

    LHS

    = ( r^2 + rh/2 + rh/2 ) / 3

    >= ∛ [(r^2)(rh/2)(rh/2)]

    = ∛ [ (r^4) (h^2) / 4 ]

    = RHS

    2010-07-23 20:07:57 補充:

    If the volume is fixed ,

    r^2 = rh/2 = rh/2

    r = h/2

    r : h = 1 : 2

    2010-07-23 20:17:14 補充:

    S = 2πr^2 + 2πrh

    S = 2πr^2 + 2 (πr^2)h / r

    S = 2πr^2 + 2 V / r

    ;;;;;;

    S / (6π) >= ∛ ( V^2 / 4π^2 )

    [2πr^2 + 2 (πr^2)h / r] / (6π) >= ∛ [ (πr^2 h)^2 / 4π^2) ]

    ( r^2 + rh ) / 3 >= ∛ [ (r^4) (h^2) / 4 ]

    LHS

    = ( r^2 + rh/2 + rh/2 ) / 3

    >= ∛ [(r^2)(rh/2)(rh/2)]

    = ∛ [ (r^4) (h^2) / 4 ]

    = RHS

    ;;;;;;;;

    ' = ' holds if and only if r^2 = rh/2

    r : h = 1 : 2 when the surface area is a minimum.

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