## Maximum Product

#### Problem

For any positive integer, $k$, let $S_k = \{x_1,x_2,...,x_n\}$ be the set of real numbers for which $x_1 + x_2 + ... + x_n = k$ and $P = x_1 x_2 ... x_n$ is maximised.

For example, when $k = 10$, the set $\{2, 3, 5\}$ would give $P = 30$ and the set $\{2.2, 2.4, 2.5, 2.9\}$ would give $P = 38.25$. In fact, $S_10 = \{2.5, 2.5, 2.5, 2.5\}$, for which $P = 39.0625$.

Prove that $P$ is maximised when all the elements of $S$ are equal in value and rational.

#### Solution

This proof will make use of the AM-GM inequality, which states that for any set of real numbers their arithmetic mean is greater than or equal to their geometric mean.

$\dfrac{x_1 + x_2 + ... + x_n}{n} \ge (x_1 x_2 ... x_n)^{1/n}$

In particular, equality is given if and only if $x_1 = x_2 = ... = x_n$.

$\therefore \dfrac{x_1 + x_2 + ... + x_n}{n} = \dfrac{k}{n} \ge (x_1 x_2 ... x_n)^{1/n}$
$\therefore \left ( \dfrac{k}{n} \right )^n \ge x_1 x_2 ... x_n = P$

As $P \le \left( \dfrac{k}{n} \right )^n$ we demonstrate that $P$ will be maximised when the terms are all equal.

We now consider the maximum value of $P = \left ( \dfrac{k}{n} \right )^n = e^{n ln{ \left ( \frac{k}{n} \right ) } }$.

$\therefore P' = e^{n ln{ \left ( \frac{k}{n} \right ) } }(1.ln{ \left ( \frac{k}{n} \right ) } - 1) = \left ( \dfrac{k}{n} \right )^n (ln{ \left ( \frac{k}{n} \right ) } - 1)^n$

Solving $P' = 0$ we get $ln{ \left ( \frac{k}{n} \right ) } = 1 \implies \dfrac{k}{n} = e \implies n = \dfrac{k}{e}$.

As $n$ represents the number of elements in $S$ it is necessary for $n$ to be integer. That is, $P$ will be maximised when $n = \left [ \dfrac{k}{n} \right]$ or $n = \left [ \dfrac{k}{n} \right]+1$ where $[ \ ]$ is the integer part function.

Whichever produces the maximum, $x_1 = x_2 = ... = x_n = \dfrac{k}{n}$ will be rational.

Problem ID: 334 (19 Nov 2007)     Difficulty: 4 Star

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