jamieg

f(x,y)=(x+1)² - (y-1)² + x(y-2)

The first row really gives the first term away--only y^x could produce 1s across the board at x=0 and non-trivial values after. Then the fact that it's and odd function (or whatever you'd call a multivariable function with rotational symmetry over y=x) gives

f(x,y)=y^x-x^y

 

 

Start by assuming the final rectangle must have integer values. The total area of the mentioned rectangles is 1056 (source). The possible rectangles with integer values with area 1056 will be the pairings of the lowest factor of 1056 with its highest factor, or its second-lowest with its second-highest, or third-lowest/third-highest, etc:

 

(1, 1056); (2, 528); (3, 352); (4, 264); (6, 176); (8, 132); (11, 96); (12, 88); (16, 66); (22, 48); (24, 44); (32, 33). (source)

 

We can rule out any with width<18 because it couldn't possibly hold our largest square, leaving us with just the final 3:

 

(22, 48); (24, 44); (32, 33).

 

 

We would have to rule out any with a length less than 15+18=33, but all three pass this test. 

 

We then have to assume for just a second that the 14-piece will be found somewhere adjacent to the 15-piece (if the 18 piece is in the bottom left and the 15 piece in the bottom right, the most efficient place for the 14-piece in order to keep a small border would be above the 15 piece). This means the width must be at least 29, ruling out (22,48) and (24,44), leaving us with just (32,33). 

 

Edited: (22,48) is possible by putting the three largest pieces lengthwise as 15+18+14=47, but this would obviously lead to problems when trying to fill up to hit the 48th unit lengthwise as we have only one 1-piece.

 

Now that we have the shape of the final rectangle, trial and error (with some common sense to speed things up)gives us the final shape. Click here for final.

 

 

 

 

Maximum is 48 drops with 2 bulbs:
drop at floor 48. If bulb breaks, increments of 1 from floor 1 to floor 47 (max 48 if breaks at floor 47). If bulb survives first drop, go to floor 48+47=95 (2nd drop). If break, increments of 1 from 49 to 94 (max 48 if breaks at floor 93). If bulb survives at 95, go to floor 48+47+46=141. If breaks at 141 (third drop), incremental from 96 to 140 (max 48 if breaks at floor 140).
Note that the pattern will get you to 48+47+46+45+...+3+2+1, which is overkill (sum=1035), but starting at 47 won't get you to 1000 floors. So 48 is our magic number.

As far as extrapolating for future steps, my instinct would be to do a similar procedure for n=3, where, if the bulb breaks (eg) at 48, then you go to floor 10, then if break incremental; if not, floor 10+9=19, etc etc, which will have a max of 10 within the step, for a max of 11; however, 11 isn't our maximum because we know it can take 48 drops to get to the top floors. So there's some other way that takes these steps to make for even distribution regardless of the floor, e.g. floor 14^2, then floor 14^2+13^2, then 14^2+13^2+12^2, etc, and then a pattern within that following the n=2 system (N+(N-1)+(N-2)...+1). But I'll let someone else do the actual optimization and give all the glory to him/her/them.