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?
AM>=GM
2010-07-24 2:49 am
回答 (1)
2010-07-24 4:17 am
✔ 最佳答案
∛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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