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

Explanation

Let the radius of base be 'r', height be 'h' and slant height be 'l' of the cone.
Its total surface area
` S= pi r^(2) + pi r l `
` rArr l = (S-pi r^(2))/(pi r)`
Volume of cone ` V = 1/3 pi r^(2) h`
` rArr V^(2)= 1/9 pi ^(2)r^(4)h^(2)`
` V^(2)= 1/9 pi^(2) r^(4) (l^(2)-r^(2))`
` = 1/9 pi^(2)r^(4)[((S-pir^(2))/(pi r))^(2)-r^(2)]`
`=1/9 pi^(2)r^(4)[((S-pir^(2))^(2)-pi^(2)r^(4))/(pi^(2)r^(2))]`
`1/9 r^(2)[S^(2)-2piSr^(2)]`
` Z = V^(2) = S/9 [Sr^(2)-2pi r^(4)]`
`(dZ)/(dr) = S/9 [2S r - 8 pi r^(3) ] `
Now ` (d^(2)Z)/(dr^(2) = S/9 (2S - 24 pi r^(2))`
For maxima/minima ` (dZ)/(dr) = 0`
` rArr 2Sr = 8 pi r^(3)`
` rArr 2(pi r^(2) + pi r l) = 8 pi r^(2)`
` rArr pi r l = 3 pi r^(2)`
` rArr l = 3r rArr r/l = 1/3`
` rArr sin alpha = 1/3 " where " alpha` = semi-vertical angle.
` rArr alpha = sin ^(-1) 1/3`
at ` l = 3r`
` (d^(2)Z)/(dr^(2) )= S/9 (2S - 24 pi r^(2))`
` (2S)/9 (pi r^(2)+ pi r l - 12 pi r^(2))`
` = pi r* (2S)/9 (l - 11 r)`
` = (2S pi r)/9 (-8 r) lt 0`
Therefore, Z is maximum.
` rArr V^(2)` is maximum.
` rArr ` V is maximum.
Therefore, for maximum volume, semi-vertical angle ` = sin ^(-1) 1/3`.