Some sort of geometry thing that is nevertheless a calc prob

[scrum]

A right circular cylinder is inscribed in a sphere of radius r. Find the dimensions of such a cylinder with the largest possible volume (your answer may depend on r).

I'm currently trying to find the angles of an inscribed cube but i haven't done that yet. this is supposedly an optimization calc problem but i'm trying to just take an inscribed cube and then using a sidelength as the height and going from there.


base radius =

height =

 

[tkhunny]

Notice a few things:

I'm thinking of a 2D drawing, more or less a circle with a rectangle in it.

1) A diagonal of the cylinder is a diameter of the sphere. Call "R" the radius of the sphere.

2) Call the angle from one cylinder corner, to the center, then to the other corner on the same side, \(\displaystyle 2\alpha\).

3) Call the radius of the cylinder, "r". Observe that \(\displaystyle \cos(\alpha)\;=\;\frac{r}{R}\).

4) Call the height of the cylinder, "h". Observe that \(\displaystyle \sin(\alpha)\;=\;\frac{h/2}{R}\).

5) \(\displaystyle r^{2}\;+\;\left(\frac{h}{2}\right)^{2}\;=\;R^{2}\)

There's a ton of stuff to think about.

 

[Dr. Flim-Flam]

Volume of a cylinder = Pi*R^2*h.

Hence looking at the great circle of the sphere, we have r^2=R^2+h^2/4, r being the radius of the circle.

R^2=r^2-h^2/4. V(cyl)= Pi*[r^2-h^2/4]h. dV/dh = PI*r^2-(3/4)*Pi*h^2. Setting the derivative = to 0, we have

h=2r/(? 3). Ergo R = ?( 2/3) r.

Hence Max volume of cylinder = (4?r^3)/(3?3), r being the radius of the sphere.