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

Explanation

Let `r,h,l` be the radius, height and slant height of the right circular cone respectively.
Let `s` be the given surface area of the cone.
we have, `l^2=r^2+h^2`........(1)
`s=pirl+pir^2`
`s-pir^2=pirl`
=>`=(s-pir^2)/(pir)`........(2)
`v=1/3pir^2h`
=>`v=1/3pir^2sqrt(l^2-r^2)` (by(1))
=>`v^2=1/9pi^2r^4(l^2-r^2)`
=>`v^2=1/9pi^2r^4[((s-pir^2)/(pir))^2-r^2]`
=>`v^2=1/9pi^2r^4[((s-pir^2)^2-(pi^2r^4))/(pir)^2]`
=>`v^2=1/9r^2[(s-pir^2)^2-pi^2r^4]`
=>`v^2=1/9r^2[s^2-2pisr^2+pi^2r^4-pi^2r^4]`
=>`v^2=1/9(r^2s^2-2pisr^4)`
`2v(dv)/(dr)=s^2/(9)2r-(2pis)/9(4r^3)`
`2v(dv)/(dr)=(2rs)/9(s-4pir^2)`
for maximum volume,`(dv)/(dr)=0`
=>`(2rs)/9(s-4pir^2)=0`
=>`2rs=0=>r=0` or `s-4pir^2=0`
since, `r` cannot be `0`
=>`s=4pir^2`
on simplification
=>`r^2=s/(4pi)`
`r^2=(pirl+pir^2)/(4pi)`
=>`4pir^2=pirl+pir^2`
=>`l=3r`
let `alpha` be the semi-vertical angle
`sinalpha=r/l`
`sinalpha=r/(3r)`
=>`alpha=sin^(-1)(1/3)`