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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)?

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    $\begingroup$ It is not clear from your wording what constraints are given and what freedoms are allowed. For example, for k=1 and a much smaller than b, I can use a square with side length close to 0.7 times b, if I get to choose the packing. Do I get to choose? Or are the orientations fixed and I have to find the smallest square covering them? Gerhard "What Are The Rules Here?" Paseman, 2018.08.31. $\endgroup$ – Gerhard Paseman Aug 31 '18 at 22:50
  • $\begingroup$ What motivates this question? And why are the cases $k=4$ and $k=6$ of special interest? $\endgroup$ – Wlodek Kuperberg Nov 6 '18 at 21:50
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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.

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