jmc

Let $BC=40x$ and $AC=41x.$ By Triangle Inequality, $x$ must satisfy $$\begin{array}{rl}9+40x& >41x,\\ 9+41x& >40x,\\ 40x+41x& >9.\end{array}$$The first inequality tells us $x<9,$ the second inequality always holds, and the third inequality tells us $x>\frac{1}{9}.$

The semi-perimeter is $s=\frac{9+81x}{2},$ so by Heron's formula, $$\begin{array}{rl}[ABC{]}^2& =\frac{9+81x}{2}\cdot \frac{81x-9}{2}\cdot \frac{9+x}{2}\cdot \frac{9-x}{2}\\ & =\frac{81}{16}(9x+1)(9x-1)(9+x)(9-x)\\ & =\frac{81}{16}(81x^2-1)(81-x^2)\\ & =\frac{1}{16}(81x^2-1)(81^2-81x^2).\end{array}$$By AM-GM, $$(81x^2-1)(81^2-81x^2)\le {\left[\frac{(81x^2-1)+(81^2-81x^2)}{2}\right]}^2=3280^2,$$so $$[ABC]\le \sqrt{\frac{3280^2}{16}}=820.$$Equality occurs when $81x^2-1=81^2-81x^2,$ or $x^2=\frac{3281}{81},$ so the maximum area is $\boxed{820}.$