Hello, Philip!

2. Determine the dimensions of the cone of maximum volume

that can be inscribed in a sphere with radius 3 Code:

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

The center of the sphere is $\displaystyle O.$

Note that $\displaystyle OA = OB = OC = 3.$

$\displaystyle \Delta ABC$ is the side view of the cone.

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

In the diagram, the radius is $\displaystyle x$ and the height is $\displaystyle y + 3$

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

The diagram also says: .$\displaystyle x^2 + y^2 \:=\:3^2\quad\Rightarrow\quad x^2 \:=\:9 - y^2$ **[2]**

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

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

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

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 the measurements are positive, $\displaystyle y = 1$

. . and from **[2]**: .$\displaystyle x^2 \:=\:9-1^2\:=\:8\quad\Rightarrow\quad x \,= \,2\sqrt{2}$

Therefore, the cone of maximum volume has radius $\displaystyle 2\sqrt{2}$ and height $\displaystyle 4.$