When multiplying 2 positive integers, why when we round the larger number up and the smaller number down, the product will be lower and vice versa. I experimented with 2 digits multiplication and found out that when we rounded the larger number up and the smaller number down, the product will be lower. When we rounded the larger number down and the smaller number up, the product will be higher.
For example (We first multiply 97 by 84):
$97\times84=8148$
$98\times83=8134$
$99\times82=8118$  
$96\times85=8160$
$95\times86=8170$
$94\times87=8178$ 
$90\times90=8100$
$90\times91=8190$  
My questions are:
1. How can we explain this? Do we need to use any inequality such as the Geometric Mean or something?
2. It seems that $90\times91=8190$ is the highest, if we we plot a graph can we explain why is it the maximum?
3. Is it a general property of integer, are there any other systems with other operations that have this property, can we explain it more abstractly?
Thank you!
 A: Compare the graphs $y=x^2$ and $y=(x-a)(x+a)$ Calculate the coordinates of their vertices. That is one approach. The other approach is to foil $(x-a)(x+a)$ and realized that we now have a parabola that is $a^2$ units shifted down. The highest y-coordinate for the vertex is the one where $a=0$. I am sure there is a way how to explain it with Geometric Mean, but this is the way how I learned it. In your example $x = 90.5$ goes the same way
A: You could think of a rectangle with a constant perimeter. Although the perimeter is constant the side length can be changed. Do you know for a constant perimeter which rectangle has a greatest area? It's a square and that's has some relation with the side length.
Now let's move to algebra. WLOG $a>b$ and let $a \cdot b = x$, then add one to $a$ and we subtract one from $b$.
$$(a+1)(b-1) = ab - a + b - 1 = x - (a+1-b)$$
Because $a>b$ it means that the second term will be bigger than zero so the value od the expresion will be smaller than the initial one.
It's easy to see that the maximum occurs when $a=b$ or $a=b+1$, depending on their parity.
A: Consider $a+b=2k$ and assume WLOG $a\ge b$
Then $a=k+x$, $b=k-x$ for any $x\ge0$
Finally, $ab=k^2-x^2$, and since
$$x^2\ge0\implies k^2-x^2\ge k^2$$
And equality is achieved when $x=0$, that is, $a=b$. Finally, we note that we note that "rounding $a$ down and $b$ up" is precisely minimizing $x$.
A: If $x$ and $y$ are integers with $x < y$, then $x + 1 \le y$, so $xy \le xy - x + y - 1 = (x + 1)(y - 1)$.  That's your basic result right there.  If $x + 1 < y$, then the same idea shows that $xy < (x + 1)(y - 1)$, so you can only get equality when $x$ and $y$ are at most $1$ apart.
