Make the following transformations:
$$mn = \left(\frac{m+n}{2}\right)^2 - \left(\frac{m-n}{2}\right)^2 = 45^2 - \left(\frac{m-(90-m)}{2}\right)^2 = 2025 - (m-45)^2 \rightarrow \max.$$
So the max will be achieved, when the value of the expression $(m - 45)^2$ is minimized, but here we have to not forget that $m, n$ are coprime.

$m = 45$, then $n = 45$ and numbers $m, n$ aren't coprime.

$m = 46$, then $n = 44$ and numbers $m, n$ aren't coprime.

$m = 47$, then $n = 43$ and numbers $m, n$ are coprime, as required. So the largest value of the product is $2025 - (47 - 45)^2 = 2021$.