Maximize the area of this triangle

Write the area of the triangle through Heron's formula as:

$A=\sqrt {s(s-a)(s-b)(s-c)}$ where $a,b,c$ are the sides of the triangle and $s$ is half of the triangle's perimeter.

After putting in the values we get:

$A=k\sqrt {(81-x^2)(81x^2-1)}$ where k is some constant.

The maximum value of the function inside of the square root occurs at some $x=\alpha$ where $[\alpha]=6$ .

[.] is the floor function.

This problem can be solved even without using calculus:

Let a=41x, b = 40x and c = 9 .

Then, the half-perimeter of the triangle:

$s = \frac {81x +9}{2} = \frac {9}{2}(9x+1)$

The area of the triangle, by using Heron's formula:

$A(x) = \sqrt { s(s-a)(s-b)(s-c) } = \sqrt { \frac {9}{2}(9x+1) × \frac {x +9}{2} × \frac {9-x}{2} ×\frac {9}{2}(9x-1) } = \frac {9}{4} \sqrt { (81x^2-1)(81-x^2) }$

Now, according to the triangle inequality, and the fact that x is an integer (which has to be positive, as 40x and 41x are side lengths):

$9 + 40x < 41x \Rightarrow 9 < x \Rightarrow 1 \leq x \leq 8$

Therefore, it is enough to calculate the areas in 8 cases only (and it is very easy to make a table e.g. by using a scientific calculator or a spreadsheet).

The relevant values (2 d. p.):

A(1) = 180 A(2) = 354.83 A(3) = 515.12 A(4) = 652.79 A(5) = 757.49 A(6) = 814.90 (maximum) A(7) = 801.75 A(8) = 667.87

Hence, our answer should be:

$x = \boxed {6}$