Do you know the formula?

There exists some $k \in \mathbb{R}$ such that $BC = 40k$ and $AC = 41k$ . Let $f(k)$ represent the area of the triangle. The semi-perimeter of the triangle is $s = \frac{9+40k+41k}{2} = \frac{9+81k}{2}$ . Therefore, using Heron's Formula, we get:

$f(k) = \sqrt{s(s-9)(s-40k)(s-41k)} = \sqrt{\left(\frac{9+81k}{2}\right)\left(\frac{-9+81k}{2}\right)\left(\frac{9+k}{2}\right)\left(\frac{9-k}{2}\right)}$ $f(k) = \frac{9}{4} \sqrt{(81k^2-1)(81-k^2)}$

We are trying to find the maximum, so let's find where the derivative is equal to zero:

$f'(k) = \frac{9}{4} * \frac{1}{2 \sqrt{(81k^2-1)(81-k^2)}} * \left(162k(81-k^2) - 2k(81k^2 - 1) \right) = 0$ $162k(81-k^2) - 2k(81k^2 - 1) = 0$ $81(81 - k^2) - (81k^2 - 1) = 0$ $6562 = 162k^2$ $k^2 = 3281 / 81$

If we plug this into the formula for the area, we get $f(\sqrt{\frac{3281}{81}}) = \boxed{820}$ .

I suppose I should also prove that this is a maximum instead of a minimum. We can see this by verifying that $f(\frac{1}{9}) = f(9) = 0$ . Therefore $f$ is increasing on $(1/9, \sqrt{3281 / 81})$ and decreasing on $(\sqrt{3281 / 81}, 9)$ . Therefore, this is indeed a maximum.

Given parameter of a triangle, maximum area is that of a right angled triangle.
This is a problem where a side is given and other two sides are also given with an unknow n, so the sides are 40n, 41n.
9, 40n, 41n is a right angled triangle when n=1.
So the area is 1/2 * 40 * 41=820.