Calculus : Maximum and Minimum

Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius  R .

Solution

$\displaystyle \text{Let}\ \text{the}\ \text{volume}\ \text{of}\ \text{cylinder}\ =\ V=\pi {{r}^{2}}h$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ {{r}^{2}}+{{\left( {\frac{h}{2}} \right)}^{2}}={{R}^{2}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ {{r}^{2}}={{R}^{2}}-\frac{{{{h}^{2}}}}{4}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ V=\pi \left( {{{R}^{2}}-\frac{{{{h}^{2}}}}{4}} \right)h$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\pi \left( {{{R}^{2}}h-\frac{1}{4}{{h}^{3}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \frac{{dV}}{{dh}}=\pi \left( {{{R}^{2}}-\frac{3}{4}{{h}^{2}}} \right)$ 

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

$\displaystyle \ \ \ \ \ \ \ \pi \left( {{{R}^{2}}-\frac{3}{4}{{h}^{2}}} \right)=0$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ {{R}^{2}}-\frac{3}{4}{{h}^{2}}=0$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{h}^{2}}=\frac{{4{{R}^{2}}}}{3}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h=\frac{{2R}}{{\sqrt{3}}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{h}^{2}}}}=-\frac{3}{2}\pi h$ 

$\displaystyle \text{When}\ \ h=\frac{{2R}}{{\sqrt{3}}}\ \ ,\ \ \ \ \frac{{{{d}^{2}}V}}{{d{{h}^{2}}}}=-\sqrt{3}\ \pi R\ <\ 0$ 

$\displaystyle V\ \text{is}\ \text{maximum}\ \text{value}\ \text{when}\ \ h=\frac{{2R}}{{\sqrt{3}}}\ $ 

$\displaystyle \ \ \ \ \ \ \ {{r}^{2}}={{R}^{2}}-\frac{{{{R}^{2}}}}{3}=\frac{{2{{R}^{2}}}}{3}$ 

$\displaystyle \therefore \ \ \ r\ =\sqrt{{\frac{2}{3}}}\ R$