For the 4/7 problem, split two of the muffins into 1/3s, and the other two into 3 * 5/21 + 2/7. Six of the parties get 1/3 + 5/21, and the last gets 2/7 + 2/7. Minimum size is 5/21, > 3/14. Rich ------------ Quoting James Propp <jpropp@cs.uml.edu>:
Alan Frank (a friend of mine and Andy Latto's) asks, what's the best way to divide m identical muffins among n people, so that each person gets the same amount of muffin (receiving either one piece of size m/n or several pieces whose sizes add up to m/n), where "best" means "so as to maximize the size of the smallest piece".
For example, if 4 muffins are to be divided among 7 people, and if each piece has size of the form i/7 for some i, then it is easy to show that one of the pieces must have size 1/7 (it isn't possible for everyone to get either 4/7 or 2/7+2/7). But we can improve on this by dividing two of the muffins 5:5:4 and dividing the other two 3:3:8 and then giving everyone either 5/14+3/14, 4/14+4/14, or 8/14 of a muffin; this is better since the smallest piece has size 3/14 > 1/7.
Jim Propp
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun