Tiling a rectangle with the maximum number of squares of side length greater than $s$

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I am 3D modelling a small parts storage organizer when I came across the idea of having variously sized cubic compartments tiling a rectangular drawer. A thought this would make for an interesting design, and would allow me to store a variety of differently-sized parts. At this point, the problem of actually determining how to tile the rectangular drawer struck me. One obvious constraint is that each square compartment should be greater than or equal to some minimum size (I'm storing screws, gears, etc. not atoms). This leads me to the following two questions:

  1. Given a rectangle of length $l$ and width $w$, is it necessarily possible to tile said rectangle (that is, leaving no unused space) with squares whose side lengths are all greater than or equal to $s$, where $s \leq \min(w, l)$
  2. If so, is there an algorithmic approach to finding the ideal arrangement (the one with the most square compartments) for a given $l$, $w$, and $s$. Otherwise, when is it definitively not possible?

Note: the side lengths of each square need not be integral.