# A cylinder is inscribed in a right circular cone of height 6 and radius (at the base) equal to 5. What are the dimensions of such a cylinder which has maximum volume?

May 5, 2015

The inscribed cylinder's volume will be maximized when its radius is $r = \frac{10}{3} u n i t s$ and its height is $2$ $u n i t s$.

Let $r$ be the radius of the cylinder inscribed in the cone and let $y$ be the cylinder's height. The volume of the cylinder is,

$V = B \cdot h = \left(\setminus \pi {r}^{2}\right) \cdot y$

Place an edge of the cone at the origin and take the $x$ to be the distance to the edge of the inscribed cylinder from the outside of the cone. The diagram below shows $x$.

From the diagram, the radius of the inscribed cylinder is $r = 5 - x$. We can also express the height of the cylinder, $y$, in terms of $x$ because we know the slope of the slanted side of the cone. For every five units we move to the right we go up 6 units.

Slope$=$ $\frac{r i s e}{r u n} = \frac{6}{5}$

$\setminus \implies y = \frac{6}{5} x$

Substitute $r$ and $y$ into the top expression for $V$

$V = \setminus \pi {\left(5 - x\right)}^{2} \cdot \frac{6}{5} x = \frac{6}{5} \pi \left({x}^{3} - 10 {x}^{2} + 25 x\right)$

To find the maximum value of $V$, find when $\frac{d}{\mathrm{dx}} V = 0$.

Keep in mind that $0 \setminus \le q x \setminus \le q 5$ so that the left edge of the inscribed cylinder stays properly within the cone.

$\frac{d}{\mathrm{dx}} V = \frac{6}{5} \pi \left(3 {x}^{2} - 20 x + 25\right) = 0$

$\setminus \implies 3 {x}^{2} - 20 x + 25 = 0$

Use the quadratic formula to find $x$

$x = \frac{20 \setminus \pm \setminus \sqrt{400 - 4 \cdot 3 \cdot 25}}{2 \cdot 3} = \frac{20 \setminus \pm 10}{6}$

The answer between 0 and 5 is $x = \frac{5}{3}$

You can verify that $x = \frac{5}{3}$ gives the point of a maximum by graphing $V$ in terms of $x$.

At $x = \frac{5}{3}$,,,,, $y = \frac{6}{5} x = 2$ and $r = 5 - x = \frac{10}{3}$