A manufacturer sends out his candy tubes in boxes of 160. They are packed in eight rows of 20 each, and exactly fill the box. Could he, by packing differently, get more than 160 in the box? If so, what is the greatest number he could add? At first sight, it sounds absurd to expect to get more candy tubes into a box that is exactly filled, but a moment's consideration should give you the key to the paradox.
The Joke and Riddle people couldn't get it, maybe you guys can
2007-01-13 08:52:55 · 6 answers · asked by MM&FF 2 in Science & Mathematics ➔ Mathematics
Let
r = radius of a tube
h = height of box
If the tubes are packed in the box in 8 rows of 20 then:
h = 8*2r = 16r
To fit more tubes in the box we have to devise an alternative packing method that will fit in 9 rows.
If we stack alternate rows by staggering the placement of an upper row into the low spots between the tubes on the row beneath, then the vertical distance between the centers of the tubes from one row to the next is:
(2r)(√3/2) = (√3)r
So the height of nine rows under this scheme is:
r + 7(√3)r + r = (2 + 8√3)r = 15.9r < 16r so 9 rows will fit.
The rows stagger alternating between 20 and 19 tubes per row. So the number of tubes that will fit in the box is:
5*20 + 4*19 = 176 tubes
2007-01-13 11:53:46 · answer #1 · answered by Northstar 7 · 1⤊ 3⤋
This is a simple packing problem. When the box is filled with 8 rows of 20 tubes each, each tube touches four other tubes at the "quarters" positions, so each tube is centered exactly one tube diameter from these tubes.
If the box is packed with a row of 20 tubes and the next row is positioned between the centers of the tubes below it, it will only hold 19 tubes, but will not "fill the box" as high. The next row would contain 20 tubes, just like the first row.
If 5 rows of 20 (100) and 4 rows of 19 (76) now fill the box, the box will contain 176 tubes of candy in the same volume.
2007-01-13 09:20:53 · answer #2 · answered by Richard 7 · 14⤊ 1⤋
You can get a ninth row by arranging them in a triangular pattern, dropping one from the alternating rows. The number of candy tubes you can get this way is 20 * 5 + 19 * 4 = 176.
Also, to prove the ninth row fits, the calculation is 8 * sqrt(3) / 2 + 1 = 7.93 (rounded). This is less than 8, so you have room for the ninth row.
2007-01-13 09:06:26 · answer #3 · answered by Anonymous · 7⤊ 1⤋
If the tubes are cylindrical, he's losing the space between them by stacking them in that manner. I guess he could try stacking them in a way that's akin to the stars on the American flag, but I'm not sure if that would allow him to get more tubes in a box.
Another thing he could try is to find a rectanguloid(?) box of the same volume as the cylindrical tubes he uses and pack his candy in those. That would eliminate the space between the inner-packed containers.....though I'll bet I'm overthinking this now, huh?
2007-01-13 09:01:53 · answer #4 · answered by wheezer_april_4th_1966 7 · 0⤊ 2⤋
Can't he stand the tubes up so they would take up less space....?
2007-01-13 08:58:23 · answer #5 · answered by Anonymous · 0⤊ 1⤋
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2007-01-13 09:46:35 · answer #6 · answered by ? 3 · 0⤊ 4⤋