Using the semi-perimeter formula which states that $A=\sqrt{s(s-a)^2(s-c)}$, where $s=\frac{1}{2}(2a+c)$. We know that $2a+c=1$, therefore $c=1-2a$ and $s=\frac{1}{2}$. Substituting into the area formula we get
$A=\sqrt{\frac{1}{2}(\frac{1}{2}-a)^2(\frac{1}{2}-c)}=\sqrt{\frac{1}{2}(\frac{1}{2}-a)^2(-\frac{1}{2}+2a)}$
First notice that the area is zero for $a=1/4$ (i.e. $c=1/2$, zero height) and $a=1/2$ (i.e. $c=0$, zero width). Geometrically we cannot have $a<1/4$ because the it would violate the triangle inequality. Also, geometrically we cannot have $a> 1/2$ because we could not have a perimeter of 1 (i.e. $c$ would have to be negative). So the maximum clearly lies between the following bounds
$ 1/4 < a < 1/2.$
To find the maximum (without loss of generality) we can look for extremes of the following:
$A^2=\frac{1}{2}(\frac{1}{2}-a)^2(-\frac{1}{2}+2a)$
Taking the derivative and setting to zero yields the following after simplification:
$(\frac{1}{2}-a)(1-3a)=0.$
Which yields a maximum for $a=1/3$ (i.e. $c=1/3$). The maximum area for a fixed perimeter is an equilateral triangle, we might have guessed as much.