Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface area is cot⁻¹√2.

Volume , V=1/3πr²h — (1)
Surface area, A=πr√r²+h² — (2)
Since volume if constant, we can write h in terms of V from (1):
h=3V/πr² — (3)
Using (3) in (2)
A=πr√r²+(9V²/π²r⁴) ⇒A=√π²r⁴+9V²/r² ⇒dA/dr=(4π²r³+9V²(⁻3/r⁴))/(2√π²r⁴+9V²/r²)
Now A is maximum or minimum when dA/dr=0
So dA/dr=0 when 4π²r³=-18V²/r⁴
⇒dA/dr=0 when 4π²r⁷=-18V²
⇒dA/dr=0 when 4/18π²r⁶=1/3πr²h²
⇒dA/dr=0 when 2r⁶=r⁴h²
⇒dA/dr=0 when 2r²=h²
⇒dA/dr=0 when h/r=√2
⇒dA/dr=0 when cot θ=√2
Hence proved.