The largest right circular cone

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3 .

Solution

$ \displaystyle \text{Volume}\ \text{of}\ \text{right}\ \text{circular}\ \text{cone}\ =V=\frac{1}{3}\pi {{x}^{2}}(3+y)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}\pi {{x}^{2}}(\ 3+\sqrt{{9-{{x}^{2}}}}\ )$

$ \displaystyle \frac{{dV}}{{dx}}=\frac{1}{3}\pi \left[ {{{x}^{2}}.\frac{1}{2}{{{(9-{{x}^{2}})}}^{{-\frac{1}{2}}}}(-2x)+(\ 3+\sqrt{{9-{{x}^{2}}}}\ ).2x} \right]$

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {-\frac{{{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}+6x+2x\sqrt{{9-{{x}^{2}}}}} \right]$ 

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {6x+\frac{{-{{x}^{3}}+18x-2{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {6x+\frac{{18x-3{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$

$ \displaystyle \ \ \ \ \ =\pi \left[ {2x+\frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$ 

$ \displaystyle \text{For}\ \text{stationary}\ \text{value}\ ,\ \frac{{dV}}{{dx}}=0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \pi \left[ {2x+\frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]=0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}=-2x$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6-{{x}^{2}}=-2\sqrt{{9-{{x}^{2}}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 36-12{{x}^{2}}+{{x}^{4}}=36-4{{x}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{4}}-8{{x}^{2}}=0$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}=8$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\sqrt{8}$

$ \displaystyle \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=\pi \left[ {2+\frac{{\sqrt{{9-{{x}^{2}}}}(6-3{{x}^{2}})-(6x-{{x}^{3}})\frac{1}{2}{{{(9-{{x}^{2}})}}^{{-\frac{1}{2}}}}(-2x)}}{{9-{{x}^{2}}}}} \right]$

$ \displaystyle \ \ \ \ \ \ =\pi \left[ {2+\frac{{(9-{{x}^{2}})(6-3{{x}^{2}})+{{x}^{2}}(6-{{x}^{2}})}}{{(9-{{x}^{2}})\sqrt{{9-{{x}^{2}}}}}}} \right]$ 

$ \displaystyle \text{when}\ x=\sqrt{8}\ ,\ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=\pi (2-18-16)=-32\pi <0\ $

$ \displaystyle V\ \text{is}\ \text{maximum}\ \text{value}\ \text{when}\ x=\sqrt{8}\ .$ 

$ \displaystyle \text{The}\ \text{volume}\ \text{of}\ \text{the}\ \text{largest}\ \text{right}\ \text{circular}\ \text{cone}\ =V=\frac{1}{3}\times \frac{{22}}{7}\times 32$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{704}}{{21}}\ \text{cube}\ \text{units}$