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Good evening,

I was solving an interesting problem from IMO 1976 :

Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is 1976.

And I was wondering how could we solve it if it was a product of positive real numbers ?

Numerically I'm sure we would find something with method of Lagrange multipliers, but is it possible to find something without any computer knowing that $\displaystyle\frac{1976}{e}$ is the maximum of $f : x \longmapsto \left( \displaystyle\frac{1976}{x} \right)^x$ ?

Waiting for your ideas !

LexLarn
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1 Answers1

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As with constraint $x + y = c$, maximum for $xy$ is when $x = y = c/2$, all numbers in the product are equal (if there are two different numbers, replacing both with their average doesn't change sum and increases product).

So, each number is $\frac{1976}{n}$, and product is $\left(\frac{1976}{n}\right)^n$. Then you can use that this function is monotonic and on reals has maximum at $n = \frac{1976}{e}$ to prove that maximum on integer is either $\left\lceil \frac{1976}{e} \right\rceil$ or $\left\lfloor \frac{1976}{e} \right\rfloor$. Because for different values of $1976$ one or other can be larger, I don't think you can choose without some additional calculations.

mihaild
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