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Show that semi-vertical angle of right circular cone of given surface area and maximum volume is `sin^(-1)(1/3)`. |
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Answer» 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 = (pi r^(2) + pi rl) =` constant `:. l = ((S)/(pi r) - r)`...(i) Now, `V = (1)/(3) pi r^(2) h = (1)/(3) pi r^(2) sqrt(l^(2) - r^(2))` `:. V^(2) = (1)/(9) pi^(2) r^(4) (l^(2) - r^(2)) = (1)/(9) pi^(2) r^(4).{((S)/(pi r) - r)^(2) - r^(2)}` [using (i)] `=(1)/(9) S (Sr^(2) - 2pi r^(4))` Thus, `V^(2) = ((1)/(9) S^(2) r^(2) - (2pi)/(9) Sr^(4))` `:. 2V .(dV)/(dr) = ((2)/(9) S^(2) r - (8 p S)/(9) r^(3)) = (2rS)/(9) (S - 4pi r^(2))`...(ii) `:. (dV)/(dr) = 0 rArr r = 0 or (S - 4pi r^(2)) = 0 rArr r^(2) = (S)/(4pi)` (neglecting `r = 0`) On differentiating (ii), we get `2((dV)/(dr))^(2) + 2V.(d^(2)V)/(dr^(2)) = (1)/(9) S (2S - 24 pi r^(2))` Putting `(dV)/(dr) = 0 and r^(2) = (S)/(4pi)`, we get `2V.(d^(2)V)/(dr^(2)) = (1)/(9) S (2S - 6S) = - (4)/(9) S^(2) lt 0` `:.` when the volume is maximum, we have `r^(2) = (S)/(4pi) = ((pi r^(2) + pi rl))/(4pi) rArr l = 3r` Now if `alpha` is the semivertical angle of the cone then `(r)/(l) = sin alpha rArr (r)/(3r) = sin alpha rArr sin alpha = (1)/(3) rArr alpha = sin^(-1) ((1)/(3))` Hence, the semivertical angle of a right cone of a given surface and maximum volume is `sin^(-1) (1//3)` |
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