Show that the semi vertical angle of a right circular cone of given total surface area and maximumvolume is sin^-1 (1/3)

If the height is h and the base radius is r we have
V = ⅓πr²h … (i)
S = πr√(r²+h²) + πr² … (ii) … (S is a constant)
From (ii) (S/π−r²)² = r⁴+ r²h² → S²/π²−2Sr²/π = r²h² … (iii)
Anticipating use in (i) this gives h²r⁴ = S²r²/π²−2Sr⁴/π
Using (i) V² = (π²/9)( S²r²/π²−2Sr⁴/π ) = (S/9)( Sr²−2πr⁴ )
V is greatest when V² is greatest so we just set derivative to zero
(S/9)( 2Sr−8πr³ ) = 0 → r² = S/(4π)
Substituting in (iii) leads to h² = 2S/π
∴ tan(θ) = r/h = 1/√8
From a right-angled triangle of sides 1, √8 and 3, sin(θ) = ⅓

θ =