[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.

I'm assuming from the text that r>1
Let the x by rx rectangle always be oriented horizontally so its height is x, width is rx.
[Part 1.]  
Put the 1 by r rectangle vertically to the west, the 2 by 2r horizontally to the east.  Let x=r.  The coverage ratio is (r + 4r)/(rx^2) = (5r)/r^3 = 5/r^2.
A tight fit is achieved if xr = r^2 = 1+ 2r.
r^2 - r - 1 = 0 --> r = 1 + sqrt(2)
x = r = 1 + sqrt(2) = about 2.414
Coverage ratio = 5 / (1+sqrt(2))^2 = 5 / (3+2sqrt(2) = about 85.79%

[Part 2.]
The coverage ratio is (r + 4r + 9r)/(rx^2) = 14/x^2 but the rectangles must of course fit.  Place rectangles 2 and 3 vertically side by side.  These can be enclosed in a rectangle 5 by 3r. Let x = 3r and let rectangle x's long side = 5.
So 3r^2 = 5, r = sqrt(5/3).  x = 3r = 3*sqrt(5/3).  The small box is 1 by sqrt(5/3) and the open space in rectangle x is 2 by sqrt(5/3), so the smallest rectangle fits.
r = sqrt(5/3) = sqrt(15)/3 = about 1.291
x = 3r = sqrt(15) = about 3.873
Coverage ratio =  14/xr = 14 / (3*sqrt(5/3)*sqrt(5/3)) = 14/15 = about 93.33%

Rough schematics for orientation, parts 1 and 2:
1
1222222
1222222

33322
33322
33322
33311

Edited on February 9, 2023, 12:20 am

Posted by Larry on 2023-02-09 00:19:34