A sphere of radius is given Let R and H be the radius and the height of the cone, respectively.
The volume of the cone is given by = \(\frac{1}{3}\pi r^2h\)
Now, from right angled triangle , we have
By differentiating with respect to , we get
\(\frac{dV}{dR} = \frac{d}{dr}\)\(\bigg(\frac{1}{3} \pi R^2 r + \frac{1}{3} \pi R^2\bigg)\)
=\(\frac{1}{3}\pi r \times 2R + \frac{1}{3} \pi R^2 \times \frac{-2R}{2\sqrt{r^2 -R^2}} + \frac{1}{3} \pi \times 2R \sqrt{r^2 - R^2}\)
\((\because \frac{d}{dx}(u.v) = u (\frac{dv}{dx})+(\frac{du}{dx}). v\, and \frac{d}{dx}x^n = nx^{n-1})\)
= \(\frac{2\pi}{3}rR\) - \(\frac{\pi R^3}{3\sqrt{r^2-R^2}} +\frac{2 \pi}{3}\) R \(\sqrt{r^2 - R^2}\)
=\(\frac{2\pi}{3}r R\) \(\frac{2 \pi R(r^2 - R^2) - \pi R^3}{3 \sqrt{r^2 - R^2}} = \frac{2 \pi}{3}rR\) + \(\frac{2 \pi R r^2 - 3\pi R^3}{3\sqrt{r^2 - R}}\)
Hence, \(\frac{dV}{dR} = \frac{2\pi}{3} rR + \frac{2\pi Rr^2 - 3 \pi R^3}{3\sqrt{r^2-R^2}}\)
Now, \(\frac{dV}{dR} = 0\, gives \frac{2\pi}{3} rR\) +\(\frac{2\pi Rr^2 - 3 \pi R^3}{3\sqrt{r^2-R^2}} = 0\)
⇒\(\frac{2\pi}{3}rR\) =\(\frac{3\pi R^3 - 2\pi R r^2}{3\sqrt{r^2 -R^2}}\)
⇒ 2\(\sqrt{r^2 - R^2}\) = 3R2 − 2r2
⇒ 4r2 (r2 - R2) = (3R2 − 2r2 )2 (By squaring both sides)
⇒ 44 − 4r2R2 = 94 − 12 r2R2 + 4r4 ( \(\because\) (a + b)2 = a2 − 2ab + b2 )
⇒ 9R4 − 82R2 = 0
⇒ R2 (9R2 − 82 ) = 0
Since, if R = 0, then there is no cone, but we want maximum volume of cone, therefore R ≠ 0.
Therefore, 9R2 − 82 = 0
⇒ R2 = \(\frac{8r^2}{9}\)
Now differentiating \(\frac{dV}{dR}\) with respect to R, we get
Then, by second derivative test of maxima and minima of function, we know that the volume of cone becomes maximum only when R2 = \(\frac{8r^2}{9}\).
Therefore, when R2 = \(\frac{8r^2}{9}\), then height of the cone is h = r + \(\sqrt{r^2 + R^2}\) = r + \(\sqrt{r^2-\frac{8r^2}{9}}\)
= r + \(\sqrt{\frac{r^2}{9}}\) = r + \(\frac{r}{3}\) =\(\frac{4r}{3}\)
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4r}{3}\) . (Hence proved)