[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