For example, I believe the following arrangement solves both questions above for N=4. (I'll let you determine the distances.)
| 444444X44 1 333333 2 2
Source: Original. I've limited N to 6 to keep the problem simple to approach, but I'd be happy to accept and publish results for higher N.
Hereafter the distances that I have found according to the two cases: N case 1 case 2 1 1/2 1 2 4/3 2 3 9/4 4 4 16/5 61/10 5 191/42 17/2 6 85/14 183/16 7 61/8 1171/80 The attached JPG file 011003sol.jpg (750K) gives the corresponding diagrams. The scales of the rods have been respected.
For N = 1 or 2 there is no real choice. For N = 1 you hang the rod horizontally at its middle. Longest horizontal distance from the loop is 1/2. Longest total horizontal distance is 1. For N = 2, you have the 2 horizontally at a point so that it balances with the 1 at its end. Say it is hung X units from the end with the 1. Then (2-X)(2-X)/2 = X*X/2 + 1*X, 2 - 2X + X^2/2 = X^2/2 + X, 2 - 2X = X 2 = 3X X = 2/3. The distance from the loop is thus 4/3, and the total distance is 2. For N = 3, you maximize both distances by hanging the 3 horizontal, hanging the 1 from its end, and hanging the 2 horizontal from the bottom of the 1. The 2 must be balanced at its middle, and the 3 is balanced against the 3 units of weight hanging at its end, X units from that end. (3-X)(3-X)/2 = X*X/2 + 3X 9/2 - 3X = 3X 9/2 = 6X X = 3/4. So the 3 extends 3/4 unit toward the other rods and 9/4 units the other way. The 2 extends an additional 1 unit beyond the end of the 3, for a total of 7/4 on that side of the original loop. Thus, the max distance from the loop is 9/4 and the max total distance is 4. For N = 4, the diagram given in the puzzle is best. The balance of the 4 rod is as follows, with X units on the short end: (4-X)(4-X)/2 = X*X/2 + 6X 8 - 4X = 6X 8 = 10X X = 4/5. 4/5 unit of the rod hangs on one side, with 3 and 1/5 units protruding on the other side. Let X represent the short end of the 3 rod: (3-X)(3-X)/2 = X*X/2 + 2X 9/2 - 3X = 2X 9/2 = 5X X = 9/10. So 2 and 1/10 units of the 3 rod protrudes beyond the short end of the 4 rod. That makes 2 and 9/10 on that side of the main loop, 3 and 2/10 on the other side, and 6 and 1/10 total width. By this point I've solved the same algebra 4 times for different numbers. Let's find a general solution. For a rod of length L, with a weight W hanging from one end, the distance X from the weight where the rod is balanced is: (L-X)(L-X)/2 = X*X/2 + WX (L^2)/2 - LX = WX (L^2)/2 = (L+W)X X = (L^2)/[2(L+W)] For the case of putting the long rod at the top and hanging all the weight off one of its ends, L+W = the sum of 1 to L, or L*(L+1)/2. This reduces to X = L/(L+1) in this case, explaining the pattern which is evident in the preceding results. For 5 rods I think this is the best arrangement: | 55555555X55 1 44444444 2 2 333X333 We know now that 5/6 unit of the 5 rod will be on the short side, and thus 4 and 1/6 units on the long side. For the 4 rod, L=4 and W=5; X=16/18 = 8/9. The 3 and 1/9 units of the 4 rod which sticks out here beats the 1.5 + 8/9 on the other side of the 1, and beats the 2 + a very small bit of the 3 rod you could get from hanging the rods in the order 5 2 3 1 4. The total horizontal distance is 8 and 1/9; the longest horizontal distance is that of the 5 rod, 4 and 1/6. For 6 rods, the similar arrangement (with the 6 and 5 horizontal, and the 4 hanging as dead weight off the 3) gives 5 and 1/7 horizontal extension on the 6 rod. The short end of the 5 rod will be given by L = 5, W = 9, X = 25/28. So the 5 rod sticks out 4 and 3/28 past the short end of the 6, making the total width 10 and 3/28. However, there are several competing arrangements: | 6666666666X66 2 2 44444444 1 5555555555 3 3 3 In this arrangement, the length of the short end of the 4 rod is given by L=4, W=9, X = (L^2)/[2(L+W)] = 16/26 = 8/13. The length of the short end of the 5 rod is given by L=5, W=3, X = 25/16. This gives the total extension to the right of the 6 rod to be 8/13 + 5 - 25/16 = 4 + (8*16 - 9*13)/208 = 4 and 11/208. Not as good as the original. | 6666666666X66 2 2 44444444 3 1 3 55555X55555 3 In this case we have weights hanging from both ends of the 4. This expanded equation looks like (using Z as the weight on the long end): (L-X)(L-X)/2 + Z(L-X) = X*X/2 + WX (L^2)/2 - LX + ZL - ZX = WX (L^2)/2 + ZL = (W+L+Z)X X = L*(L/2 + Z)/(W+L+Z) So with L=4, W=6, Z=3, we get X = 20/13. The end of the 5 rod is thus 2.5 + 20/13 = 4 and 1/26 units beyond the end of the 6, still less than the first case. | 6666666666X66 2 2 333333 1 5555555555 4 4 4 4 In this arrangement, the length of the short end of the 3 rod is given by L=3, W=10, X = (L^2)/[2(L+W)] = 9/26. The length of the short end of the 5 rod is given by L=5, W=4, X = 25/18. This gives the total extension to the right of the 6 rod to be 9/26 + 5 - 25/18 = 4 + (9*18 - 7*26)/468 = 4 - 20/468 < 4. There are some other arrangements, but each is clearly inferior to one of these. So we still have 10 and 3/28 total width and 5 and 1/7 horizontal extension as the best for N=6. However, for longer N, maybe as soon as for N=7, there will be enough extra dead weight around to make some of these other arrangements useful. Eventually they should be able to pass the 2N+1 width limit that arrangements in the pattern of these solutions for 4, 5, and 6 rods are approaching.