Maximizing The Area With A Fixed Perimeter

Relevant wiki: Area of Triangles - Heron's Formula

We wish to find a relationship between $a+b+c=2s$ and $\sqrt{s(s-a)(s-b)(s-c)}$ .

By AM-GM Inequality, $(s-a)(s-b)(s-c) \leq (\dfrac{(s-a)+(s-b)+(s-c)}{3})^3=\dfrac{s^3}{27}$ which has equality when $a=b=c$ .

Hence the triangle with most area is $\boxed{\mbox{Equilateral triangle}}$

Another way to do this question (Great one to prove by the way) is as follows:

Fix the base $AB$ a certain length. The locus of $C$ , the third vertex of the triangle, is an ellipse with foci $A$ and $B$ since the perimeter is also fixed. We must also have that to maximise the area of a triangle with a fixed base, its height must be maximised. This happens when $C$ is on the perpendicular bisector of $AB$ as well, thus making $ABC$ an isosceles triangle with $AC=BC$ . We can apply this argument for the ellipse with foci $B$ and $C$ , and $C$ and $A$ to get that $AB=BC=CA$ , which implies that the triangle must be equilateral to maximise its area.