smallest square containing k non-overlapping equal rectangles at any orientation This seems like something that should have a known answer, but I haven't found it after some time alternating between searching and generating multiple pages of algebra.  I'm interested in $k=4$ and $k=6$, but I'll phrase it for general $k$:
Given $k$ identical rectangles of dimension $a \times b$ (let's assume $a \leq b$), what is the smallest square that can contain them if the rectangles do not overlap and can be placed at any orientation (i.e., not necessarily parallel to the square's sides)?
 A: You might recheck your values for k=1 and 2.  In particular, an ab rectangle fits in a square of side length (a+b)/c, where c*c=2.  For a less than (c-1)b, this improves upon a square of side length b.  For k=2 and a smaller than (c-1)b/2, a similar diagonal packing works.
One approach you can use is to consider packing rectangles into the smallest rectangle, and then fitting this larger rectangle inside a square. This may not achieve optimum, but it leverages earlier work and should give near optima.
Gerhard "And Check Out Pack-O-Mania Website" Paseman, 2018.08.31.
A: Just a note to indicate why this question might not
be straightforward, now that it's been bumped to the homepage.
For $k=2$ rectangles of size $a \times b = 1 \times 6$,
the diagonal packing is better than $6$, because $2a < (\sqrt{2}-1) b$,
as Gerhard mentions in his answer.
But this is not the optimal packing, as the figure below indicates:
Tilting the rectangles differently fits in a smaller square.

          


          

The diagonal packing for two $1 \times 6$ rectangles is suboptimal.
The blue squares are identical.


It would be interesting to find an example where the rectangles
are necessarily not parallel in an optimal packing.
