show that the semi vertical angle of a right circular cone of a given surface area and maximum volume is sin^-1 (1/3)

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Tushar Kohli answered this

Let r be the radius, l be the slant height and h be the height of the cone.

∴ Surface area of the cone (S),

Let V be the volume of the cone.

Differentiating (2) w.r.t. r, we get–

For maximum or minimum,

⇒ either r = 0 or S – 4πr ² = 0

Differentiating (3), we have –

Now if α be the semi-vertical angle of the cone, then

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Sandeep Jha answered this

ncert ka ques ha ye to

Sandeep Jha answered this

Let r be the radius, l be the slant height and h be the height of the cone of given surface area S.

Then S = πrl + πr ²

$n n \Rightarrow n l n = n \frac{n}{n S n n n - n π n n^{n r n n n 2 n n} n n n} n ��.......... (1) n n$

Let V be the volume of the cone.

$n n Then �V n n n = n \frac{n}{n 1 n} n π n n^{n r n n n 2 n n} n h n n n n$

$n n \Rightarrow n n^{n V n n n 2 n n} n = n \frac{n}{n 1 n} n n^{n n n π n n n 2 n n} n n^{n n n r n n n 4 n n} n n^{n n n h n n n 2 n n} n = n n \frac{n}{n 1 n} n n^{n n n π n n n 2 n n} n n n n^{n r n n n 4 n n} n n (n n^{n n l n n n 2 n n} n - n n n^{n r n n n 2 n n} n) n n n n n \Rightarrow n n^{n V n n n 2 n n} n = n \frac{n}{n n^{n π n n n 2 n n} n} n n n n^{n r n n n 4 n n} n n n n [n n (n \frac{n}{n n S n n n n - n n n π n n n n n n n^{n r n n n 2 n n} n} n n n^{n) n n n 2 n n} n n - n n^{n r n n n 2 n n} n] n n n n n n \Rightarrow n n^{n V n n n 2 n n} n = n \frac{n}{n n^{n π n n n 2 n n} n n n n n^{n r n n n 4 n n} n} n n n n n [n n \frac{n}{n (n S n n n - n n n π n n n n^{n r n n n 2 n n} n n n n^{n) n n n 2 n n} n - n n^{n π n n n 2 n n} n n n n^{n r n n n 4 n n} n} n n] n n n n n n n n n n n n n \Rightarrow n n^{n V n n n 2 n n} n = n \frac{n}{n 1 n} n n n n n^{n r n n n 2 n n} n n n \frac{n}{n (n n^{n n n S n n n 2 n n} n n n - n 2 n n^{n n n S π r n n n 2 n n} n) n} n n n n n n \Rightarrow n n^{n V n n n 2 n n} n = n \frac{n}{n 1 n} n n n n S n n^{n r n n n 2 n n} n n n (n S n - n 2 n n^{n n π r n n n 2 n n} n) n n n n n n n n n n n n n n n n$

Let Y = V² Then V is maximum on minimum according to Z is maximum or minimum.

$n n ∴ n n n Y n n n = n \frac{n}{n 1 n} n S n n n n^{n r n n n 2 n n} n n (n S n n n - n 2 n π n n^{n r n n n 2 n n} n n n) n n n n n n n n = n \frac{n}{n 1 n} n S n n (n S n n n n^{n r n n n 2 n n} n - n 2 n π n n^{n r n n n 4 n n} n n n) n n n n$

Differentiating w rt r

$n n \Rightarrow n \frac{n}{n d n Y n} n = n \frac{n}{n 1 n} n S n n n n (n 2.5 n r n n n - n 8 n π n n^{n r n n n 3 n n} n n n) n n n n ��........... �(2) n n$

For maximum or minimum, put $n n \frac{n}{n d n n n Y n n n} n = n 0 n n$

$n n \Rightarrow n \frac{n}{n 1 n} n S n n n n (n 2 n S n n n r n n n - n 8 n π n n^{n r n n n 3 n n} n) n = n 0 n n n n \Rightarrow n 2 n S n n n r n = n 8 n π n n^{n r n n n 3 n n} n \Rightarrow n n n S n n n = n 4 n π n n^{n r n n n 2 n n} n n n n n n n n n n n n n$

Differentiating (2) wrt r,

$n n \frac{n}{n n^{n d n n n 2 n n} n n n Y n} n = n \frac{n}{n S n} n n n (n 2 n n n S n n n - n 24 n n n n π n n n^{n r n n n 2 n n} n n n) n n n n n \Rightarrow n \frac{n}{n n^{n d n n n 2 n n} n Y n} n n n n n = n n n \frac{n}{n S n} n n (n 25 n n n - n n n 24 n π� n \frac{n}{n S n} n n n n n) n n n n n n n n n, n n n (n p n u n t n \begin{matrix} n \\ n & n n n^{n r n n n 2 n n} n = n \frac{n}{n S n} n & n \\ n \end{matrix} n) n n n n n = n - n \frac{n}{n 4 n n^{n S n n n 2 n n} n} n < n 0 n n n n n n$

Thus, Y is maximum when S= 4πr ²

⇒ V is maximum when S = 4πr ²

Now, S = 4πr ²

⇒πrl + πr ² = 4πr ²

⇒πrl = 3πr ²

⇒l = 3r

Consider, in right ∆OBC, $n n \sin n n n α n = n \frac{n}{n OB n} n = n \frac{n}{n r n} n n = n n \frac{n}{n r n} n = n \frac{n}{n 1 n} n n$

$n n n n \Rightarrow n α n = n n^{n \sin n n n - n 1 n n n n n} n n (n \frac{n}{n 1 n} n) n n n n n$

Thus, V is maximum when $n n n n α n = n n^{n \sin n n n - n 1 n n} n n (n \frac{n}{n 1 n} n) n n n n n$

Sandeep Jha answered this

this may b help ful

show that the semi vertical angle of a right circular cone of a given surface area and maximum volume is sin -1 (1/3)

show that the semi vertical angle of a right circular cone of a given surface area and maximum volume is sin^-1 (1/3)