You have N rectangles (N > 1). The 2N numbers used by the length of short and long edges of these rectangles are all different positive integers.
You can create a larger rectangle by using all of these rectangles. The large rectangle is fully covered without any overlap or overflow of the smaller rectangles. What is the minimum possible area of the large rectangle?
Note: The large rectangle cannot be a square.
I think it may be impossible. (I could well be wrong.) I can almost see a proof: Two rectangles of different heights, side by side, have a gap at either their top or bottom or both. Adding yet another rectangle to fill a gap must produce yet a new gap. This is true for the sides aligned or the bases and tops aligned. Since new gaps are always made: there can be no closure to make a uniform rectangle. Perhaps....
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A comment added later, after the solution by B. Smith:
Surely his is the right answer since the integers are all minimal.
My logical error was not seeing that the continuing "gap" can circle and the last added rectangle fills the first and last gap.
Kudos, B. Smith!
Edited on January 11, 2019, 1:55 am
a guess
