Question

A right circular cylinder is inscribed in a cone with height \(h\) and base radius \(r\). Find the largest possible volume of such a cylinder.

Short answer

The largest possible volume of the cylinder is, \(V\left( {\frac{{2r}}{3}} \right) = \frac{{4\pi h{r^2}}}{{27}}\).

Step-by-step solution

To sketch the figure and mention given data

Let\(h\)be the height of the cone and\(y\)be the height of the cylinder.

Let\(r\)be the radius of the base of the cone and\(x\)be the radius of the cylinder.

The sketch is shown in the figure below:

 To find the largest possible volume of the cylinder

Since the triangle are similar, we have:

\(\begin{aligned}{c}\frac{{BC}}{{DE}} &= \frac{{AB}}{{AD}}\\\frac{x}{r} &= \frac{{h - y}}{h}\\\frac{x}{r} &= 1 - \frac{y}{h}\\y &= h\left( {1 - \frac{x}{r}} \right)\end{aligned}\)

Volume of cylinder is:

\(\begin{aligned}{c}V &= \pi {x^2}y\\V &= \pi {x^2}\left( {h\left( {1 - \frac{x}{r}} \right)} \right)\\V\left( x \right) &= \pi h\left( {{x^2} - \frac{{{x^3}}}{r}} \right)\,\,\, \cdots \cdots \left( 1 \right)\end{aligned}\)

Differentiating equation\(\left( 1 \right)\)we get,

\(V'\left( x \right) = \pi h\left( {2x - \frac{{3{x^2}}}{r}} \right)\)

Solving\(V'\left( x \right) = 0\),we get,

\(\begin{aligned}{c}\pi h\left( {2x - \frac{{3{x^2}}}{r}} \right) &= 0\\2x - \frac{{3{x^2}}}{r} &= 0\\x\left( {2 - \frac{{3x}}{r}} \right) &= 0\end{aligned}\)

For volume to be maximum,\(x\)cannot be zero.

Therefore,

\(\begin{aligned}{c}\left( {2 - \frac{{3x}}{r}} \right) &= 0\\\frac{{3x}}{r} &= 2\\x &= \frac{{2r}}{3}\end{aligned}\)

 To find the largest possible volume of the cylinder

Substituting this value in\(\left( 1 \right)\), we get,

\(\begin{aligned}{c}V\left( {\frac{{2r}}{3}} \right) &= \pi h\left( {{{\left( {\frac{{2r}}{3}} \right)}^2} - \frac{{{{\left( {\frac{{2r}}{3}} \right)}^3}}}{r}} \right)\\V\left( {\frac{{2r}}{3}} \right) &= \pi h\left( {{{\frac{{4r}}{9}}^2} - \frac{{8{r^2}}}{{27}}} \right)\\V\left( {\frac{{2r}}{3}} \right) &= \pi h\left( {\frac{{12{r^2} - 8{r^2}}}{{27}}} \right)\\V\left( {\frac{{2r}}{3}} \right) &= \frac{{4\pi h{r^2}}}{{27}}\end{aligned\)

Hence, the largest possible volume of the cylinder is, \(V\left( {\frac{{2r}}{3}} \right) = \frac{{4\pi h{r^2}}}{{27}}\).