The beginnings of a solution, where warmer colors correspond to areas, and cooler colors to perimeters (Last updated 6-26):
1, 2, 7, 15, and 47 are solved as in Michael's answer above.
6 must now be the perimeter of a 1x2 domino, otherwise we reach the paradox alluded to in Michael's answer. 29 is forced into the horizontal position, in turn forcing 23 horizontal and making 8 the perimeter of a 1x3 block. 12 must fill the blank spot to the right of 29, with height either 5 or 12. 19 must be horizontal.
17 must fill the corner between 7 and 15, because if 46 (as a 2x21 block) fills it, then 17 must fill the corner between it and 15, and 19 must fill the corner between it and 29, but then a blank spot to the left of 12 cannot be filled. 9 must then be a 3x3 block.
4 cannot be taller than 1 block, because then the corner between 9 and 12 cannot be filled without blocking the corner between 12 and the edge. 4 then must be the perimeter of a 1x1 block because otherwise 27 is forced to fill a gap of width-2, an impossibility. 37 fills the corner since 42 cannot, forcing 16 to be the perimeter of a 5x3. Various other forcings yield the diagram.
I believe that also
5 and 35 are forced to form a contiguous rectangle due to 32's position, forcing 46 to be the perimeter of an 11x12.