Hello, lauren72!

Did you make a sketch?

A cylinder is inscribed in a right circular cone of height 4 and base radius 4.5.

What are the dimensions of such a cylinder which has maximum volume? Look at a cross-section of the cone. Code:

- *
: * | *
: * | *
: * | *
4 * - - - + - - - *
: * | | | *
: * | |h | *
: * | | | *
- *-------+-------+-------+-------*
: 4.5 : r : 4.5-r :

The radius of the cylinder is $\displaystyle r$, its height is $\displaystyle h.$

From the similar right triangles: .$\displaystyle \frac{h}{\frac{9}{2}-r} \:=\:\frac{4}{\frac{9}{2}} \quad\Rightarrow\quad h \:=\:\frac{4}{9}(9-2r)$ .[1]

The volume of a cylinder is: .$\displaystyle V \;=\;\pi r^2 h$ .[2]

Substitute [1] into [2]: .$\displaystyle V \;=\;\pi r^2\cdot\frac{4}{9}(9-2r) \quad\Rightarrow\quad V \;=\;\frac{4\pi}{9}(9r^2-2r^3) $

Differentiate and equate to zero: .$\displaystyle V' \;=\;\frac{4\pi}{9}(18r - 6r^2) \;=\;0$

Then: .$\displaystyle 18r-6r^3 \:=\:0 \quad\Rightarrow\quad 6r(3-r) \:=\:0 $

Hence: .$\displaystyle \begin{Bmatrix}r \:=\:0 & \text{minimum volume} \\ r \:=\:3 & \text{maximum volume} \end{Bmatrix}\quad\Rightarrow\quad r \,=\,3$

Substitute into [1]: .$\displaystyle h \:=\:\frac{4}{9}(9-2\!\cdot\!3) \:=\:\frac{4}{3}$

Therefore: .$\displaystyle r = 3,\;h = \frac{4}{3}$