How to find the minimum product of two or more quantities when their sum is given?

Question

Suppose we have been given $ \sum_1^n x_i = S $

How will we find $x_i $ such that $ \prod_1^n x_i $ is minimum?

Any hints are appreciated. Thanks!

Edit : Sorry, my mistake. All $x_i$ are positive.

Answer 1

Does not exist.

$\prod\limits_{k=1}^nx_k$ can be less than all positive number, but $\prod\limits_{k=1}^nx_k>0$.

Try $x_1=x_2=...=x_{n-1}\rightarrow0^+$ and $x_n\rightarrow S$.

Answer 2

I think you meant finding $x_i$ such that the product is maximum, not minimum, since for a minimum, just take one of the $x_i$ to be zero. Also, the problem makes sense only if you restrict to non-negative $x_i$.

So for some fixed $S$ we are trying to find $x_i$ such that $\sum x_i = S$ and $\prod x_i$ is as large as it can be.

There is a theorem that the geometric mean of a set of numbers $GM(\{x_i\})$ is less than or equal to the arithmetic mean $AM(\{x_i\})$; it is sometimes called the AM-GM theorem. The geometric mean is the $n$-th root of the product of the numbers: $GM(\{x_i\})=\left(\prod x_i)\right)^{1/n}$.

Then $$ \left(\prod x_i)\right)^{1/n} \leq \frac1n \sum x_i = \frac{S}n \\ \prod x_i \leq \left( \frac{S}n \right)^n $$ When each of the $x_i = \frac{S}n$, this bound is exactly achieved. Therefore, the desired $\{x_i\}$ happens when they are all equal to $\frac{S}n$.

Answer 3

Mark Fischler does the general case, but it may be instructive just to consider the case of $n=2$. Let the numbers be $a,b$. We are given $S=a+b$ and asked to choose $a,b$ so the product is maximized. The product $P=ab=a(S-a)=Sa-a^2$. We can take the derivative and set to zero, getting $$\frac {dP}{da}=\frac d{da}(Sa-a^2)=S-2a$$ If we set this to zero, we see $a=\frac S2$ and the second derivative test shows this is a maximum.

Now that we have it for two numbers, informally we should make any pair the same. If we have $n$ numbers with two different, we can increase the product by changing each to the average of the two. This needs some work to be a proof, but it should convince you.

Answer 4

There is no minimum of the products, but the infimum of the products is $0$.

Proof. By assumption all products are positive. Let an $\epsilon\in\> ]0,S[\>$ be given, and put $$x_1:=\epsilon,\qquad x_k:={S-\epsilon\over n-1}\quad(2\leq k\leq n)\ .$$ Then $\sum_{k=1}^n x_k=\epsilon+(n-1){S-\epsilon\over n-1}=S$, and $$0<\prod_{k=1}^n x_k=\epsilon\left({S-\epsilon\over n-1}\right)^{n-1}\leq\epsilon\left({S\over n-1}\right)^n\ .$$ As $n$ and $S$ are fixed the RHS of $(1)$ can be made arbitrarily small by choosing $\epsilon$ small enough. This proves that $$\inf\prod_{k=1}^n x_k=0\ .$$