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From: greg@harvard.UUCP (Greg)
Newsgroups: net.puzzle
Subject: Re: Find a set of positive real numbers whose sum is 100 and whose product is as large as possible.
Message-ID: <710@harvard.UUCP>
Date: Thu, 13-Feb-86 15:09:29 EST
Article-I.D.: harvard.710
Posted: Thu Feb 13 15:09:29 1986
Date-Received: Sat, 15-Feb-86 03:12:17 EST
References: <1057@decwrl.DEC.COM> <1692@bbncc5.UUCP>
Reply-To: greg@harvard.UUCP (Greg)
Organization: Harvard
Lines: 17

In article <1692@bbncc5.UUCP> larry@bbncc5.UUCP (Larry Denenberg) writes:
>If S were not a set but a multiset, this
>process could stop only when all the elements in S were equal.  If we look at
>the function x**(100/x) we find (not surprisingly) that x = e gives the
>global maximum.  So S should have 100/e copies of e.  But 100/e is not an
>integer;  it's between 36 and 37.  The best multiset is thus either
>36 copies of 100/36 or 37 copies of 100/37;  probably the latter, since
>100/e is 36.78 or so.

Fixing a number n, we are choosing the multiset {100/n, 100/n, ..., 100/n}
and taking the product of the members, which is (100/n)^n.  If we wish to
maximize this, we may simply maximize its logarithm, which is
n*(log(100)-log(n)).  Well, 36*(log(100)-log(36))=36.779445, and
37*(log(100)-log(37))=36.787334, so the second one wins.  Funny that the
maximum of x*(log(100)-log(x)) should occur when x = x*(log(100)-log(x)).
-- 
gregregreg