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Firstly, I'm not sure if this is valid or not, but so far it's been giving me the desired result, so if this theory is invalid, please let me know the exceptions.

To get the maximum result from the multiplication of n variables, $$ v_1 v_2 \cdots v_n $$ Under the constraint that all variables to be greater than 0 $$ \{v_1,v_2,\ldots,v_n \mid v_1,v_2,\ldots,v_n > 0 \} $$ the variables should equal to each other or at least close to each other $$ v_1 = v_2 = \ldots = v_n $$ For example, under constraint that: $$ a + b = 25 \implies \{a,b \mid a,b > 0 \} $$ The largest result from multiplication would be $$ 12.5 \times 12.5 $$ Another constraint example: $$ x + y + z = 8 \implies \{x,y,z \mid x,y,z > 0\} $$ The largest result from multiplication would be $$ 8/3 \times 8/3 \times 8/3 $$ Thanks!

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  • $\begingroup$ how is this calculus? $\endgroup$
    – user148697
    May 22, 2014 at 8:25
  • $\begingroup$ If the only constraint is that the variables be positive, they don't have to be close together, they just have to be as big as possible. You could set one equal to $10^100$ and the rest equal to $1$. @jonnytan999 It's optimization. $\endgroup$
    – Jack M
    May 22, 2014 at 8:26
  • $\begingroup$ See also math.stackexchange.com/questions/581849/…. $\endgroup$
    – lhf
    May 22, 2014 at 10:12

2 Answers 2

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If the sum is fixed then the arithmetic mean is fixed, and by the AM–GM inequality, the geometric mean cannot exceed the arithmetic mean, so the product cannot exceed the $n$th power of the arithmetic mean.

If all the values are the same, i.e. equal to the arithmetic mean, then the product is equal to the $n$th power of the arithmetic mean.

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  • $\begingroup$ Ah, the AM-GM inequality. So that's what it's called. Thank you! $\endgroup$ May 22, 2014 at 8:29
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Although the case of equality in the AM-GM inequality solves the problem, I propose a particular solution for two variables $x$, $y$ under the constraint $x+y+k$.

Consider the level sets $\{(x,y) \mid xy = c\}$, where $c >0$. These sets are hyperbola, symmetric with respect to the line $y=x$. Now, the constraint is the line $x+y=k$, which is orthogonal to $y=x$. By Lagrange's multiplier rule, at a maximum point $(x_0,y_0)$ of $xy$ under the given constraint, the level set must be orthogonal to the constraint. This can happen only for $x_0=y_0$, with a value of $c$ such that the hyperbola $xy=c$ is tangent to $x+y=k$ at $(x_0,y_0)$.

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