Find highest product of natural numbers which sum to $S$? Given a number $S$, how can we find the highest $P$ such that there exist natural numbers $a$, $b$, & $c$ 
where $a + b + c = S$ and $a \times b \times c = P$?
 A: A simpler problem, maximizing the product of two natural numbers whose sum is fixed, can be used to solve this (involving a product of three summands).
That is, in the case where two numbers have a (positive) fixed sum, their product is maximized by making them as nearly equal as possible.  Indeed, if $u+v = S$, then the maximum product $uv$ is obtained by taking $u = \lfloor S/2 \rfloor$ and $v = \lceil S/2 \rceil$.
Suppose that now $a+b+c = S$, where $a,b,c$ are natural numbers (zero included for the sake of simplicity).  Then should any pair of $a,b,c$ differ by more than one, we can increase the product $abc$ by adjusting that pair to be more nearly equal!  This leaves the sum fixed but increases the product (unless the product is forced to be zero because $S \lt 3$).
It follows that $abc$ is maximized (for fixed sum $a+b+c=S$) when no two of the summands differ by more than one.  Depending on the residue of $S$ mod $3$, this could mean:
(1) If $S \equiv 0 \bmod 3$, then take $a=b=c=S/3$.
(2) If $S \equiv 1 \bmod 3$, then take two of $a,b,c$ to be $\lfloor S/3 \rfloor$ and the third summand to be $\lceil S/3 \rceil$.
(3) If $S \equiv 2 \bmod 3$, then take two of $a,b,c$ to be $\lceil S/3 \rceil$ and the third summand to be $\lfloor S/3 \rfloor$.
