Question

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1(1/3) .

Answer 1

let r be the radius,
l the slant height and h the height of the cone.

let S denote the surface area and V the volume of the cone .

then S = (πr² + πrl) = constant

l = (S/πr - r)-------( 1 )

now,

V = 1/3πr²h = 1/3πr²√(l² - r²).

V² = 1/9π²r⁴(l² - r²)

V² = 1/9π²r⁴[(S/πr - r)² - r²] ----- from ( 1 )

V² = 1/9S(Sr² - 2πr⁴).

thus V² = (S²r²/9 - 2πSr⁴/9)

2V•dV/dr = (2S²r/9 - 8πS*r³/9) = 2rS/9(S - 4πr²)----( 2 )

now, dV/dr = 0
r = 0 or (S - 4πr²) = 0

r² = S/4π [neglecting r = 0].

on differentiating (2) we get,

2(dV/dr)² + 2V*d²V/dr² = 1/9S(2S - 24πr²).

putting dV/dr = 0 and r² = S/4π, we get

2V*d²V/dr² = 1/9*S(2S - 6S) = -4/9S² < 0.

when the volume is maximum, we have
r² = S/4π = (πr² + πrl)/4π

l = 3r.

now, if $\bold{\alpha}$ is semivertical angle of the cone then,
r/l = sin$\bold{\alpha}$

r/3r = sin$\bold{\alpha}$

sin$\bold{\alpha}$ = 1/3

$\bold{\alpha}$ =
$\bold{sin^{-1}(1/3)}$.
hence,the semi Vertical angle of a right cone of a given surface and maximum value is $\bold{sin^{-1}(1/3)}$.