Answer
$l/4$
Work Step by Step
Let $\mathrm{ABCD}$ be a parallelogram with perimeter $l=p$.
From the constraint we know $b=\frac{1}{2}-a$. Using this in the area function:
\[
A(a, \theta)=a(l / 2-2 a) \sin \theta=A
\]
Using the first derivative test, we can determine the critical points
\[
A_{a}=\frac{\partial A}{\partial a}=\frac{\partial}{\partial a}\left[\left(\frac{a l}{2}-a^{2}\right) \sin \theta\right]=\sin \theta(l / 2-2 a)
\]
\[
\begin{aligned}
A_{0} &=\frac{\partial A}{\partial \theta}=0 \Rightarrow\left(\frac{a l}{2}-a^{2}\right) \cos \theta=0 \\
& \Rightarrow \theta=\pi / 2 \text { or }(a=l / 2, / b=0)
\end{aligned}
\]
We note that for $b=0, A=0;$ hence $P(a, b, \theta)=P(l / 4, l / 4, \pi / 2)$ is the only critical point such that $A \neq 0$.
Using the second derivative test
\[
\begin{array}{r}
D=A_{a a} A_{l b}-\left.A_{o b}\right|_{(U / 4, l / 4, \pi / 2)}=2\left(f^{2} / 8-P^{2} / 16\right)>0 \text { and } \\
\left.A_{a n}\right|_{(0 / 4 / 4 / 4, \pi / 2)}=-2 \sin (\pi / 2)=-2<0
\end{array}
\]
This parallelogram is a square with sides $a=l / 4$ and $\ A=(l / 4)^{2}=l^{2} / 16$