Finding the triangle with the maximum area with a given perimeter

The Heron's formula for the triangle is

$$A =\sqrt{s(s-a)(s-b)(s-c)}$$

where $s=\frac p2$. Then, Apply the AM-GM inequality to get

$$A =\sqrt{s(s-a)(s-b)(s-c)} \le \left[ s \left( \frac{3s-(a+b+c)}{3} \right) ^3\right]^{1/2} = \frac{s^2}{3\sqrt3}=\frac{p^2}{12\sqrt3}$$

where the equality, or the maximum area, occurs at $a=b=c=\frac p3$.

Just for giving another way to deduce a solution. Take a closed string and leaving a fixed side, say $a$, of the triangle draw an ellipse as usual. It is evident that the triangle with the largest area occurs with an isosceles triangle because it has the same base as all but has a greater height (actually a vertical semiaxis of the ellipse).

Now the area of this isosceles triangle is $A=\dfrac a4\sqrt{(2b)^2-a^2}$ but because of $a+2b=p$ we have a fonction of $a$, call it $x$, defined by $$A=\frac x4\sqrt{p^2-2px}$$ The derivative of A being equal to $$A'=\frac{p^2-3px}{4\sqrt{p^2-2px}}$$ we see that the maximum area is taken when $a=\dfrac p3$ then $b=\dfrac p3$ too.