Q. 238. Minimum total surface area of a cone of given volume

Question 238.
(a) Prove that the total surface area of a cone of given volume V
     is minimum when its height, h is twice its radius, r.
(b) Also prove that the minimum surface area in terms of its    
     volume is S = 3(2piV^2)^(1/3).

Answer 238.
(a)
Total surface area, S = 2πrh + 2πr^2
and volume, V = πr^2h => h = V/(2πr^2)
 => S
= 2πr * V/(2πr^2) + 2πr^2
= V/r + 2πr^2

For S to be minimum, dS/dr = 0 and d^2S/dr^2 > 0
dS/dr = 0 => - V/r^2 + 4πr = 0
=> - πr^2h/r^2 + 4πr = 0
=> h = 2r
Also, d^2S/dr^2 = 2V/r^3 + 4π > 0
=> S is minimum when h = 2r for a given V.

(b)
S = 2πrh + 2πr^2
V = πr^2h and
h = 2r

Plugging h = 2r in S and V,
S = 4πr^2 + 2πr^2 = 6πr^2 and
V = πr^2 * 2r = 2πr^3

Plugging r = (V/2π)^(1/3) in S,
S
= 6π * (V/2π)^(2/3)
= 6π * (V^2/4π^2)^(1/3)
= 3 * (8π^3)^(1/3) * (V^2/4π^2)^(1/3)
= 3 * (8π^3 * V^2/4π^2)^(1/3)
= 3 (2πV^2)^(1/3).

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