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켐컴人 이라면 풀이 가능 할듯?
추천 : 44 이름 : ****** 작성일 : 2006-08-14 22:56:54 조회수 : 287
[A] Suppose that a given positive number N is partitioned into two positive numbers a and b such that a+b = N. The product ab= a(N-a) has a maximum if a=b=N/2. To see this, let P = Na - a2. By dP/da = N - 2a = 0, we obtain a = N/2 and b = N - N/2 = N/2.

Let's look at this problem in another way. We first define the two numbers A and G such that A = (a+b)/2 and G=, and then prove that G achieves its maximum when a and b are equal. If a and b are different, A - G = (a+b)/2 - = ( - )2 / 2 > 0; if they are equal, A = G. Therefore, the maximum value G can take is A and the value is achieved when a=b.

In the case of n positive numbers a1, a2, ..., an with n being no smaller than 2 such that the sum of the numbers is equal to N, we define A=(a1+a2 + ... + an) / n and G = with respect to a1, a2, ..., an. The proof can be done by changing the numbers in such a way that A remains constant but G gets larger until all the numbers are equal, at which point A and G are equal.

Suppose that in the set { a1, a2, ..., an }, a1 is bigger than any other number in the set and a2 is smaller than any other number in the set so that we have a1 > A > a2. Now replace a1 by A and a2 by a2' where a2' = a1+a2-A which is > 0. The new set of numbers becomes {A, a2', a3, ..., an}. Notice that because A+a2' = a1+a2, the sum of the numbers in the new set is equal to that in the old set. Also, the value of A with respect to the numbers in the new set, (A+ a2' + ... + an) / n, is equal to the value of A with respect to the numbers in the old set. By some algebraic manipulation, we have that Aa2' - a1a2 = A(a1+a2-A) - a1a2 = (A-a2)(a1-A) > 0. So the product of the numbers in the new set is greater than that in the old set. This means that by the construction of the new set the value of G has increased.

Based on the above arguments, we can generate sets in sequence such that the numbers in the set generated at each stage will eventually all equal A, and A and G computed with respect to the numbers in the set will become equal. From this we can show that the product of the elements in a partition of N achieves its maximum when the elements are all equal.
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