Maximize product of numbers with a given sum

The problem is:

Find a set of positive real numbers which sum to 271 and has as large product as possible.

All numbers must be $\gt0$, otherwise the product would be zero.

A candidate could be [200, 71]. Could we do better? Yes. Let’s prove that an optimal solution must have all numbers equal.

Proof - all numbers must be equal

Assume the sequence is $X=x_1,...x_N$. Without losing generality, we can require that it is sorted such that $x_i\le x_j$ for $i<j$, because rearranging the sequence does not affect the sum nor the product.

Assume we have a valid candidate $X$. Build a new sequence $Y=x_m,x_2,...x_{N-1},x_m$ which is $X$ but with the first (smallest) and last (largest) element replaced with their mean $x_m=(x_1+x_N)/2$. The sum will be unchanged. The product will be:

$$\prod{Y}=\frac{\prod{X}}{x_1 x_N} (\frac{x_1+x_N} {2})^2$$

The product $\prod{Y}$ will be greater than $\prod{X}$ if

$$\frac{(x_1+x_N)^2} {4 x_1 x_N} \gt 1$$

$$\frac{x_1}{x_N} + 2 + \frac{x_N}{x_1} \gt 4$$

This can be simplified to

$$\frac{1}{a}+a > 2$$

which is true for positive $a$ fulfilling $a\neq 1$.

That means any candidate $X$ which has non-equal elements, can be replaced with $Y$ having a strictly higher product. Thefore, the optimal solution must have all elements equal.

The optimal value

The problem boils down to selecting the number of terms $N>0$ with equal elements such that the product

$$P=\prod{X}=\prod{\frac{271}{N}}=(\frac{271}{N})^N$$

is maximized.

$$\ln{P}=N ( \ln{271} - \ln{N})$$

The derivative of this with respect to $N$ is

$$1(\ln{271} - \ln{N}) + N \frac{1}{-N}$$

which means the maximum occurs at $$\ln{N}= \ln{271}-1$$ $$N=\frac{271}{e}\approx 99.67$$

Since $N$ must be integral, we investigate $P$ for $N=99$ and $N=100$ and find that $N=100$ gives a slightly higher answer.