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Topic: 271 (Read 2708 times) |
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rugga
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Quote:| Write 271 as the sum of positive real numbers so as to maximize their product |
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I like this one because I initially didn't suspect what
turned out to be the solution, but after some trial and
error with smaller numbers I took a wild guess which
turned out correct.
Here's some hints at a proof (copying the style of
a different post by S. Owen):
Hint 1:
The way to maximize the product of
n numbers which sum to C is to make all n numbers
equal. So the largest product is (C/n)n.
So then the question becomes what value of n
maximizes (C/n)n, where n is a positive
integer.
Hint 2:
The derivative can find the real-valued n which
maximizes the above expression. Then take the closest
integer on either side and see which gives a larger answer.
To take Dn(C/n)n, rewrite it as
Dnen ln (C/n). Then apply the chain
rule and a few steps to get
(C/n)n (ln C - ln n - 1).
Setting this equal to 0, you end up with
ln n = ln C -1. Raising both sides to the power of e:
n = Ce-1 = C/e
Answer:
For this problem C=271, so n~99.695.
It's a fair bet 271 was chosen so that n would come
out even, but check just to make sure:
(271/99)99 ~ 1.977 x 1043
(271/100)100 ~ 1.981 x 1043
Anyone have a more intuitive explanation? I'm not sure
I really understand why the math works out
the way it does.
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jk
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I looked at recursively breaking 271 (or any number) down into addends (is that the word?), Arbitrarily, pick two numbers such that:
N + M = 271
then, break down N and M:
(N0 + N1) + (M0 + M1) = 271
Our product (N0 * N1)(M0 * M1), will grow as long as each component can be advantageously broken down. So, how far do you go?
1 is of course a loser: (N-1) * 1 < N
2*2 breaks even for 4
3*2 > 5 good
3*3 > 6 better
Any addend larger than 4 can be broken down to advantage.
I'd love something more elegant, but you can empirically plug in numbers and see that:
For 6: 3*3 > 2*2*2
For 8: 3*3*2 > 2*2*2*2
Three seems to be the lowest advantageous breakdown, so I propose:
271 = 4+(89 * 3)
4*3^89 = 1.16... * 10^43
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AlexH
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Posts: 156
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Re: 271
« Reply #2 on: Aug 29th, 2002, 10:44am » |
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jk: the problem specified positive real numbers, not positive integers so rugga's solution is the correct one.
rugga: In terms of finding the solution intuitive I'll say this ... note that maximizing (C/x)x is the same (substitute y=C/x) as maximizing yC/y = (y1/y)C. Since we're dealing with numbers > 1, this is the same thing as maximizing y1/y which is a standard early problem in derivatives.
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jk
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OUCH! Yes, I misread the question.
mea maxima culpa
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rugga
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Re: 271
« Reply #4 on: Sep 1st, 2002, 2:44am » |
Quote Modify
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Quote:| rugga: In terms of finding the solution intuitive I'll say this ... note that maximizing (C/x)x is the same (substitute y=C/x) as maximizing yC/y = (y1/y)C. Since we're dealing with numbers > 1, this is the same thing as maximizing y1/y which is a standard early problem in derivatives. |
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Thanks AlexH.
After thinking about it some more here's something I came up
with. The question is about maximizing the product of a bunch
of numbers, which is the same as maximizing the sum of
their logarithms. So we really want to find the number which
has the largest log (or the most "multiplicative power") per unit:
maximize (log x) / x
Again using basic calculus, this is maximized when x = e
(doesn't matter what the base of the log is).
This helped me gain some insight. I'm not sure if it will
resonate with others, but I hope someone might find it useful.
BTW, I see that maximizing (log x) / x is the same as
maximizing y1/y as you suggest. My formula
is the log of your formula.
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NickH
Senior Riddler
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Re: 271
« Reply #5 on: Sep 1st, 2002, 11:53am » |
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Another approach is to note that we seek the least integer x such that (271/x)^x > (271/(x+1))^(x+1). Simplifying, we
seek the least x such that x*(1 + 1/x)^(x+1) > 271.
We know that x is reasonably large, in which case (1 + 1/x)^(x+1) ~= e. So we seek the least x such that x > 271/e. This gives an intuitive feel as to why the terms (2.71, in this case) are close to e. It can be made more robust, if required.
Nick
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