Question

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is.

Answer 1

Let θ be the semi-vertical angle of the cone.

It is clear that

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

The slant height of the cone is given as constant.

Now, r = l sin θ and h = l cos θ

The volume (V) of the cone is given by,

$$\begin{gathered} V=\frac{1}{3}\pi r^{2}h \\ =\frac{1}{3}\pi \left(l^2{\sin }^2\theta \right)\left(l\cos \theta \right) \\ =\frac{1}{3}\mathrm{\pi }{l}^3{\sin }^2\theta \cos \theta \\ \Rightarrow \frac{dV}{d\theta }=\frac{l^3\mathrm{\pi }}{3}\left[{\sin }^2\theta \left(-\sin \theta \right)+\cos \theta \left(2\sin \theta \cos \theta \right)\right] \\ =\frac{l^3\mathrm{\pi }}{3}\left[-{\sin }^3\theta +2\sin \theta {\cos }^2\theta \right] \\ \Rightarrow \frac{d^{2}V}{d{\theta }^2}=\frac{l^3\mathrm{\pi }}{3}\left[-3{\sin }^2\theta \cos \theta +2{\cos }^3\theta -4{\sin }^2\theta \cos \theta \right] \\ =\frac{l^3\mathrm{\pi }}{3}\left[2{\cos }^3\theta -7{\sin }^2\theta \cos \theta \right] \end{gathered}$$

∴By second derivative test, the volume (V) is the maximum when.

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is.