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प्रश्न
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is `Sin^(-1) (1/3).`
उत्तर
The branch is considered to have radius, oblique, and total surface area S and volume V.
Entire page: `S = pir (r + I) or pirI = S - pir^2`
or `l = (S - pir^2)/(pir) = S/(pir) - r` ...(1)
and volume V = `1/3 pir^2h`
or `V^2 = 1/9 pi^2 r^4 h^2 = 1/9 pi^2 r^4 (l^2 - r^2)` ...[∵ Δ from OAC, h2 = l2 - r2]
or `V^2 = (pi^2 r^4)/9 [(S/(pir) - r)^2 - r^2]`
`= (pi^2 r^4)/9 [S^2/(pi^2r^2) - (2S)/pi + r^2 - r^2]`
`= pi^2/9 [(S^2 r^2)/pi^2 - (2Sr^4)/pi]`
`therefore V^2 = (S^2 r^2)/9 - (2piSr^4)/9 = u` (Let) ...(2)
Differentiating equation (2) with respect to r, `(du)/(dr) = S^2/9* 2r - 2/9 piS * 4r^3` ...(3)
For maximum or minimum value of u i.e. V2, `(du)/(dr) = 0`
i.e, `S^2/9 * 2r - 2/9 pi * S * 4r^3 = 0`
or `(2Sr)/9 [S - 4pir^2] = 0 therefore S = 4pir^2`
or `pir (l + r) = 4pir^2` or l + r = 4r
or l = 3r or `= l/3`
Differentiating equation (3) with respect to r, `(d^2u)/(dr^2) = (2S^2)/9 - 8/9 piS * 3r^2`
`S = 4pir^2 on, (d^2u)/(dr^2) = (2 (4 pir^2)^2)/9 - 8/9 pi * 4pir^2 * 3r^2`
`= (32pi^2 r^4)/9 - (96 pi^2r^4)/9 = (64pi^2 r^4)/9` (negative)
`therefore at r = l/3` there will be maximum, i.e. volume V of the cone will be maximum.
But when `r = l/3`
Then if the half apex angle of the cone is `theta`, then
`sin theta = r/l = r/(3r) = 1/3` or `theta = sin^-1 (1/3)`
Therefore, the volume of the cone will be maximum if the semi-vertex angle is `sin^-1 (1/3)`.
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