This is in continuation of
Two circles.
In a 8.5x11 sheet of paper I drew two nonoverlapping ellipses of equal area, with major axis of each equalling twice the minor axis  both completely inside the paper, of course.
What's the largest portion of the paper I could cover with these ellipses?
What would be the answer if I drew THREE equal ellipses?
The idea is to find a solution with two circles in a rectangle that is only half as wide (or half as high) and then just stretch it out.
This only works if the optimal size has ellipses with parallel major axes to one side of the rectangle.
A 4.25 by 11 rectangle is too thin for the circles to fill very well, so I will only work with a 8.5 by 5.5.
As with the two circles in the linked problem we set the circles in opposite corners and set up the quadratic:
(5.52r)^2 + (8.52r)^2 = (2r)^2
solution
r = 7.25sqrt(374) = 2.165
total area
2*pi*r^2 = (144.75  7sqrt(374))*pi = 29.457
divided by the rectangle of area 8.5*5.5 = 46.75
gives a proportion .6301
The stretching will not affect this answer.
It may be possible to tilt these ellipses to make them bigger. That will not be easy.
For 3 ellipses there is probably a solution where one of the ellipses is perpendicular to the other two.

Posted by Jer
on 20110119 13:10:54 