Show that the semi vertical angle of a right circular cone of given surface area and maximum volume is sin−11√3
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, l2=r2+h2 ....(1)
S=πrl+πr2
S−πr2=πrl
⇒l=S−πr2πr .....(2)
V=13πr2h
⇒V=13πr2√l2−r2 (by (1))
⇒V2=19π2r4(l2−r2)
⇒V2=19π2r4[(S−πr2πr)2−r2]
⇒V2=19π2r4[(S−πr2)2−π2r4π2r2]
⇒V2=19r2[(S−πr2)2−π2r4]
⇒V2=19r2[S2−2πSr2+π2r4−π2r4]
⇒V2=19(r2S2−2πSr4)
2VdVdr=S292r−2πS94r3
2VdVdr=2rS9(S−4πr2)
For maximum volume, dVdr=0
⇒2rS9(S−4πr2)=0
⇒r=0 or S−4πr2=0
Since, r cannot be 0
⇒S=4πr2
⇒r2=S4π
⇒r2=πrl+πr24π
⇒4πr2=πrl+πr2
⇒3πr2=πrl
⇒l=3r
Let α be the semi-vertical angle .
sinα=rl
sinα=r3r
⇒α=sin−1(13)