Number $N$ of summands adding up to $5$ whose product is maximal. In how many parts $N$ should the integer $5$ be partitioned so that the product of these parts is maximized?
I tried using A.M$\geq$G.M but could not go further.
 A: Suppose you partition a number $n$ into exactly $k$ summands $n_1 + n_2 + \cdots + n_k$. It is a routine calculation to show that the product $n_1 \cdots n_k$ is maximized when all summands are equal. In this case $n_i = \frac{n}{k}$ for all $1 \le i \le k$ and the maximum product is $\left (\dfrac{n}{k} \right)^k$.  
So, what value of $k$ maximizes $\left (\dfrac{n}{k} \right)^k$? Clearly $1 \le k \le n$. Define
$$f(x) = \left (\dfrac{n}{x} \right)^x, \quad 1 \le x \le n$$
and apply the first derivative test. Since $f(x) = e^{x(\log n - \log x)}$ you have
$$f'(x) = e^{x(\log n - \log x)} (\log n - \log x - 1) = e^{x(\log n - \log x)} \log\left( \frac{n}{xe}\right)$$ so that $f'(x) > 0$ if $xe < n$ and $f'(x) < 0$ if $xe > n$. 
It follows that $f$ attains its maximum value when $x = \dfrac ne$, and that the maximum value that $f(k)$ takes among integer values $k$ is attained either at
$k=\lfloor n/e \rfloor$ or $k = \lceil n/e \rceil$.
In case $n=5$ you have $1 < 5/e < 2$ so the product is maximized using a partition with either $1$ or $2$ summands. When $k=1$ the product is $5$ and when $k=2$ it is $\left(\dfrac 52 \right)^2 = 6.25$.
The maximum product is 6.25 using 2 (equal) parts.
