How to find the side for the biggest area? Let $x$ be one side of a rectangle and $a$ its perimeter. We know that it's area is given by:
$$
S = x\cdot\left(\frac{a}{2}-x\right).
$$
$$
S=-x^2+ax/2$$
 where a=-1, b= a/2 and c= 0
$$D=a^2/4$$
$$n=-D/4a=-a^2/4/-4=a^2/16$$
$$So, $$$$a^2/16$$ is the biggest value, right? But there are 2 questions:
1) For which value of $x$, we have the biggest area value?
2) From all the triangles that have the same perimeter which one has the biggest area? 

(1)  $S = x (a/2-x)$ is a product of two positive numbers with constant sum.  That gets maximised when the terms are both equal i.e. $x = a/2-x$ or $x=a/4$, to give a maximum of $a^2/16$.  Another way to see the same is to write it as $S = a^2/16-(x-a/4)^2$.
(2) Let $s$ be the fixed semi-perimeter, then by Heron's formula, the area is given by $A$, where
$$\frac{A^2}s = (s-a)(s-b)(s-c)$$
Again, this can be looked at as a product of three terms with a constant sum, viz. $s$, hence is maximised when all the terms are the same, i.e. $a=b=c$. Alternately, by AM-GM
$$\frac{A^2}s = (s-a)(s-b)(s-c) \le \left(\frac{(s-a)+(s-b)+(s-c)}3\right)^3=\frac{s^3}{27}$$
with equality possible only when $s-a=s-b=s-c \iff a=b=c$, i.e. when it is equilateral.
