1. Find the ellipse of smallest area which circumscribes 2 non-overlapping unit circles.

2. Find the ellipse of smallest area which circumscribes 3 non-overlapping unit circles.

(In reply to As far as I plan to get by Jer)

I didn't realize until I read Erik O.'s post that I actually solved part 1 and that part 2 is much easier.

If you want the ABSOLUTE minimum for ANY circles, make then touching (duh).  This makes d=1 in my previous post.  The expression to minimize (from [7]) is

b*sqrt(b^4/(b^2 - 1)

I still woudn't take the derivative but the approximate solution is a=2.1213, b=1.2247, area = 8.162

Part 2 is even easier as the minimal ellipse is a circle of radius 1+2��(3)/3 = 2.1547 which has area 14.5856

-Jer

Posted by Jer on 2004-11-29 19:30:44