# Thread: Finding the Maximum Area of a cylinder

1. ## Finding the Maximum Area of a cylinder

Here's the problem. Can someone work through it for me? I don't know where to start.

A cylinder is inscribed in a right circular cone of height 4 and radius (at the base) equal to 4.5. What are the dimensions of such a cylinder which has maximum volume?

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

3. You, my friend, are a saint.