We have to find n numbers such that $$ x_1 + x_2 + \cdots + x_n = k $$ $$ x_1 * x_2 * .....* x_n = maximum $$
What are the values of $x_1, x_2...x_n$ ?
Note that $x_1, x_2...x_n$ are all positive integers .
P.S : I had posted a similar question before. In that question, $x_1, x_2...x_n$ could be any positive numbers and did not necessarily have to be integers. This is the previous question : maximum product of n numbers whose sum is k
First suppose that the $x_i$ are restricted to non-negative integers. In this case we need the $x_i$ to be
all equal, or at least as close to equal as possible.
The reason is that
For any $d>0$ we have $$(t-d)(t+d) = t^2-d^2 < t\cdot t$$ and $$(t-d)(t+d+1) = t^2+t-d^2-d < t(t+1)$$ So if any two of the $x_i$ differ by 2 or more, increasing the smaller and decreasing the larger will increase their product.
So the values of $x_i$ are:
$\lfloor \frac{k}{n}\rfloor$ or $\lceil \frac{k}{n}\rceil$. There will be $k \bmod n$ of the latter.
Suppose now that the $x_i$ are allowed to be negative.
For $n=1$ or $n=2$ allowing negative values makes no difference, so this case is as above.
For $n\ge 3$ we can choose the following values for $x_i$:
The first three variables have values $x_1=T+k+4-n$, $x_2=-T$, $x_3=-1$, and the remaining $n-3$ variables all have value $1$. Here $T$ is an arbitrary value large enough to make $x_1$ positive. Their sum is $k$ as required, but their product can be made arbitrarily large by choosing $T$ larger.
The following answer was given by @GarethMcCaughan .
There is always an optimal answer with no two factors differing by more than 1, because if we have factors a,b with a>b+1 then replacing them with a−1,b+1 leaves the sum unaltered and yields a product that has increased by a−b−1≥0.
So
consider an optimal configuration with no two xj differing by more than 1. Letting x be the smallest of our factors, some of our n factors are x and the rest are x+1; let's say m of them are x+1 where 0≤m<n. (We can't have m=n because then the smallest number would actually be x+1.) Then the sum (which must equal k) is nx+m, so m=k mod n, and there is exactly one m with this property and 0≤m<n, namely the remainder on dividing k by n. So this configuration is unique (up to permutation of the xs).