I don't know what it's called, so it's hard to explain, but say we have the number $12$, which can be $1 \times 12$, $2 \times 6$, or $3 \times 4$.
I want the $[3, 4]$ pair because $3$ is the greatest out of the lower of each pair.
Background
The reason is that I want a faster way to calculate primes. There's no point in this case testing $1...12$ to see if there are any divisors, because there is a pair $[1,12]$ so I would have already ruled out $12$ by checking $1$. Likewise, there's no point testing up to $6$, because I would have already ruled it out by testing $2$. (Does that make sense?) Of course, I stop checking if I find any divisor, but for large primes I don't want to search all the way up to $n$ where $n$ is the number I'm testing, I should be able to say "I've gone up to $m$, and I know since there are no divisors up to $m$, there won't be any between $m$ and $n$." I want a way to calculate $m$.
We can confine trial division to numbers $\le \sqrt{n}$. For if $ab=n$, with $a$ and $b$ positive, we cannot have $a$ and $b$ both greater than $\sqrt{n}$.
The case $n=p^2$, where $p$ is prime, shows that in general we cannot do better.