Maximizing the product

If there is a 5 in the partition, we could replace it with $2+3$ , which would be a better partition since $2\times3=6$ which is greater than 5. Similarly, if there is a 6 in the partition, we could replace it with $2+4$ , where $2\times4=8$ . In fact, this is true for any number $n\geq4$ : $n\geq4\implies 2(n-2)\geq n$ We also want to avoid 1s, since they contribute nothing to the product. This means there is a partition with a maximal product containing only 2s and 3s.

So, let's start with a partition containing only 2s and 3s, and try to improve it. If we replace $2+2+2$ with $3+3$ (which give the same sum and so don't affect the rest of the partition), we replace a product of $2\times2\times2=8$ with $3\times3=9$ , increasing the overall product. Hence, we want to replace as many 2s with 3s as possible. The best partition is then $25=3+3+3+3+3+3+3+2+2=7(3)+2(2)$ with a product of $3^72^2=8748$

Fixing the number of partitions $2 \leq p \leq 12$ , each partition $x_i>1, i=1,\dots p$ . We have $\sum x_i = 25$ . If partitions dod not have to be integers, one can show that all partitions $x_i$ should be equal for the product $\prod x_i$ to be maximum. The positive integer version would be to have $p-r(25,p)$ of the partitions to be $\lfloor \frac{25}{p} \rfloor$ and $r(25,p)$ of the partitions to be $\lfloor \frac{25}{p} \rfloor+1$ . $r(x,y)$ is the remainder of dividing $x$ by $y$ . The maximum product, for a fixed $p$ is then

$\lfloor \frac{25}{p} \rfloor^{p-r(25,p)} \times (\lfloor \frac{25}{p} \rfloor+1)^{r(25,p)}$

then we can make a table, for different $2 \leq p \leq 12$ to find out that for $p=8,9$ the maximum $3^7\times 2^2=8748$ is achived.