# Cylinder inscribed in sphere

1. Oct 16, 2006

### Mindscrape

The problem is to find the radius and height of the open right circular cylinder of largest surface area that can be inscribed in a sphere of radius a. What is the largest surface area?

The open cylinder's surface area will be
$$f(h,r) = 2 \pi r h$$

I am not really sure about the sphere, because I'm not really sure about the constraints that would apply. It looks like it would just be
$$a^2 = r^2 + h^2 + r^2 = 2r^2 +h^2$$, but this would only be if the cylinder is centered about the origin. But I guess that since it is a sphere, and perfectly symmetrical, then trying to squeeze the cylinder in diagonally would be the same as along the origin. Right?

2. Oct 16, 2006

Yes, it is the same. Combine the constraint into the function and try to differentiate.

3. Oct 16, 2006

### Mindscrape

Huh, I got something funny:

$$\nabla f = \lambda \nabla g$$
$$\nabla f = (2 \pi r)i + (2 \pi h)j$$
$$\nabla g = (4r)i + (2h)j$$
so
$$2 \pi r = \lambda 4r$$
$$2 \pi h = \lambda 2h$$
which means
$$\lambda = \frac{ \pi}{2} = \pi$$???

Last edited: Oct 16, 2006
4. Oct 16, 2006

### HallsofIvy

No, I'm pretty sure $\pi\ne \frac{\pi}{2}$!

5. Oct 16, 2006

I think the constraint you set is wrong. It should be $$a^2 = r^2 + \left(\frac{h}{2}\right)^2$$, unless I'm missing something.
Ah crap, you are right. I just realized that I built the constraint wrong. But this constraint doesn't work either. The only one that would work would be $$a^2 = r^2 + h^2$$.