[Part 1.] Pack two similar rectangles with aspect ratio r and short sides of length 1 and 2 into a third similar rectangle with short side x, such that the maximum proportion of the larger rectangle is covered.
Find r, x and the proportion covered.
[Part 2.] Pack three similar rectangles with aspect ratio r and short sides of length 1, 2, and 3 into a fourth similar rectangle with short side x, such that the maximum proportion of the larger rectangle is covered.
Find r, x and the proportion covered.
Part 1: I took the 2 by 2r rectangle and put it horizontally, and put the 1 by r vertically next to it.
The bounding rectangle around that configuration is 2r+1 by r. So then this is the x*r by x rectangle, therefore we can create the equation r = (2r+1)/r.
This equation simplifies to r^2-2r-1=0, which has one positive real root at r=1+sqrt(2). Then x=1+sqrt(2).
The areas of the three rectangles are 1+sqrt(2), 4+4sqrt(2), and 7+5sqrt(2). The portion covered then is (5+5sqrt(2))/(7+5sqrt(2))=15-10sqrt(2)=0.8579
Part 2: I accidentally found this r=2 result just by sketching out cases. Place the 1 by 2 atop the 3 by 6, both horizontally; and then place the 2 by 4 vertically. The bounding box is 4 by 8, so x=4.
The areas of the rectangles are 2, 8, 18, and 32. The portion covered is (2+8+18)/32=7/8 = 0.875
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