Maximize the product of two integer variables with given sum Let's say we have a value $K$. I want to find the value where
$A+B = K$ (all $A,B,K$ are integers) where $AB$ is the highest possible value.
I've found out that it's:


*

*$a = K/2$; $b = K/2$; when $K$ is even 

*$a = (K-1)/2$; $b = ((K-1)/2)+1$; when $K$ is odd.
But is there a way to prove this?
 A: In the even case, assuming $A\leq B$, the numbers will be of the form $A=\frac{K}{2}-m$ and $B=\frac{K}{2}+m$ in order to get $A+B=K$, with $m=0,1,\ldots,\frac{K}{2}$.  Then multiplying gives $AB=\left(\frac{K}{2}\right)^2 - m^2 \leq \left(\frac{K}{2}\right)^2$.  
In the odd case, you'll have $A=\frac{K}{2}-m-\frac{1}{2}$ and $B=\frac{K}{2}+m+\frac{1}{2}$, with $m=0,1,\ldots,\frac{K-1}{2}$.  Multiplying gives 
$\begin{align*}
AB &= \left(\frac{K}{2}\right)^2 - \left(m+\frac{1}{2}\right)^2 = \frac{K^2}{4} - m^2 - m - \frac{1}{4} \\
   & = \frac{K-1}{2}\left(\frac{K-1}{2}+1\right) - m^2 - m \leq \frac{K-1}{2}\left(\frac{K-1}{2}+1\right).
\end{align*}$
Alternatively, you can think of optimizing a quadratic function by finding the vertex.  If you write $AB$ as a quadratic function of $A$ or $B$ alone using the constraint $A+B=K$ (as indicated in PEV's answer), then the closest integer to the vertex will solve your problem because the parabola opens down.
A: We know that $B = K-A$. So you want to maximize $AB = A(K-A)$. So we have:
$$AB = A(K-A)$$
$$ = AK-A^2$$
Find the critical points and use some derivative tests.
