# Using the Derivative to Analyze Polynomial Function Model Problems

• Oct 31st 2006, 12:55 PM
philipsach1
Using the Derivative to Analyze Polynomial Function Model Problems

Determine the dimensions of the cone of maximum volume that can be inscribed in a sphere with radius 3?
• Oct 31st 2006, 01:53 PM
Soroban
Hello, philipsach1!

Did you make a sketch?

Quote:

[I] Determine the dimensions of the cone of maximum volume
that can be inscribed in a sphere with radius 3?

Code:

                A               * * *           *    /:\    *         *    / : \    *       *    /  :  \    *             /  :3  \       *    /    :    \    *       *  /    O*    \  *       *  /  3*  :  *3  \  *         / *    y:    * \     B *- - - - + - - - -* C         *      :  x  *           *    :    *               * * *

The center of the sphere is $\displaystyle O.$
$\displaystyle \Delta ABC$ is the side view of the cone.
Note that: $\displaystyle OA = OB = OC = 3.$

The volume of a cone is: .$\displaystyle V \;= \;\frac{1}{3}\pi r^2h$ [1]

In the diagram, we see that: $\displaystyle x =\text{ radius, and }y + 3 =\text{ height}$

So we have: .$\displaystyle V \;= \;\frac{1}{3}\pi x^2(y + 3)$ [2]

The diagram also tells us that: .$\displaystyle x^2 + y^2\:=\:3^2\quad\Rightarrow\quad x^2 \,= \,9 - y^2$

Substitute into [2]: .$\displaystyle V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)$

Can you finish up?

• Oct 31st 2006, 05:50 PM
philipsach1
Thanks
Thank you so much! This really helped a lot.

It would be sooooo great if you could finish it up for me, I'm having so much trouble. Please and thank you.
• Nov 1st 2006, 03:37 PM
Soroban
Hello again, philipsach1!

The rest is "simple" Calculus . . . but okay.

We have: .$\displaystyle V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)$

How do we maximize the volume?
We equate its derivative to zero and solve.

The function is: .$\displaystyle V \;= \;\frac{\pi}{3}\left(-y^3 - 3y^2 + 9y + 27\right)$

So: .$\displaystyle V' \;= \;\frac{\pi}{3}\left(-3y^2 - 6y + 9\right) \;=\;-\pi(y^2 + 2y - 3) \;= \;0$

And we have the quadratic: .$\displaystyle y^2 + 2y - 3 \;= \;0$

. . which factors: .$\displaystyle (y - 1)(y + 3) \;= \;0$

. . and has roots: .$\displaystyle y \:= \:1,\;-3$

Since we want positive dimensions: .$\displaystyle y \:=\:1\quad\Rightarrow\quad x = 2\sqrt{2}$

. . $\displaystyle \begin{array}{ccc}\text{The radius is: } r \:= \:x \:= \:2\sqrt{2} \\ \\ \text{The height is: }h \:= \:y + 3 \:= \:4\end{array}$

• Nov 1st 2006, 05:51 PM
philipsach1
Thank you
Thanks so much! I really appreciate this.