Question

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac{8}{27}\) of the volume of the sphere.

Answer 1

Let \(r\) and h be the radius and height of the cone respectively inscribed in a sphere of

radius R.

Let \(V\) be the volume of the cone.

Then\(,V=\frac{1}{3}\pi r^{2}h\)

Height of the cone is given by,

\(h=R+AB=R+\sqrt{R^{2}-r^{2}} [ABC \space is\space  a\space  right\space triangle]\)

\(∴V=\frac{1}{3}\pi r^{2}h(R+\sqrt{R^{2}-r^{2})}\)

\(=\frac{1}{3}\pi r^{2}hR+\frac{1}{3}\pi r^{2}\sqrt{R^{2}-r^{2}}\)

\(∴\frac{dV}{dr}=\)\(\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{R^{2}-r^{2}}+\frac{1}{3}\pi r^{2}.\frac{(-2r)}{2\sqrt{R^{2}-r^{2}}}\)

\(=\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{R^{2}-r^{2}}-\frac{1}{3}\pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}}\)

\(=\frac{2}{3}\pi rR+\frac{2\pi r(R^{2}-r^{2})-\pi r^{3}}{3\sqrt{R^{2}-r^{2}}}\)

\(=\frac{2}{3}\pi rR+\frac{2\pi rR^{2}-3\pi r^{3}}{3\sqrt{R^{2}-r^{2}}}\)

\(\frac{d^{2}V}{dr^{2}}\)=\(\frac{2\pi R}{3}\)+\(\frac{3\sqrt{R^{2}-r^{2}}(2\pi R^{2}-9\pi r^{2})-(2\pi rR^{2}-3\pi r^{3}).\frac{(-2r)}{{6\sqrt{R^{2}-r^{2}}}}}{9(R^{2}-r^{2})}\)

=\(\frac{2}{3}\pi R\)\(+\frac{9(R^{2}-r^{2})(2\pi R^{2}-9\pi r^{2})+2\pi r^{2}+3\pi r^{4}}{27(R^{2}-r^{2})\frac{3}{2}}\)

Now,\(\frac{dV}{dr}\)\(=0⇒\)\(\pi 23rR\)=\(\frac{3\pi r^{3}-2\pi rR^{2}}{3\sqrt{R^{2}-r^{2}}}\)

\(⇒2R={3r^{2}-2R2}{\sqrt{R^{2}-r^{2}}}\)\(⇒2R\sqrt{R^{2}-r^{2}}=3r^{2}-2R^{2}\)

\(⇒4R^{2}(R^{2}-r^{2})=(3r^{2}-2R^{2})^{2}\)

\(⇒4R^{4}-4R^{2}r^{2}=9r^{4}+4R^{4}-12r^{2}R^{2}\)

\(⇒9r^{4}=8R^{2}r^{2}\)

\(⇒r^{2}=\frac{8}{9R^{2}}\)

When \(r^{2}=\frac{8}{9R^{2}},\)then\(\frac{d^{2}V}{dr^{2}}<0.\)

⧠By second derivative test,the volume of the cone is the maximum when \(r^{2}=\frac{8}{9R^{2}}.\)

When \(r^{2}=\frac{8}{9R^{2}}\),\(h=R+\sqrt{R^{2}-\frac{8}{9}R^{2}}\)\(=R+\sqrt{\frac{1}{9R^{2}}}=R+\frac{R}{3}=\frac{4}{3R.}\)

Therefore,

\(=\frac{1}{3}\pi (\frac{8}{9}R^{2})(\frac{4}{3}R)\)

\(=\frac{8}{27}(\frac{4}{3}\pi R^{3})\)

\(=\frac{8}{27}\times\)(Volume of sphere)

Hence,the volume of the largest cone that can be inscribed in the sphere is\(\frac{8}{27}\)

the volume of the sphere.