Answer
$a=b=\ell / 4$
$\ell^{2} / 16=A$
Work Step by Step
We need to maximize the area of a paralelogram $A=a b$ sin $\theta$ subject to the condition
\[
l=2 a+2 b \quad \text { perimeter }
\]
Using Lagrange Multipliers, the objective function is $A=a b \sin \theta$ subject to the constraint $g(a, b)=2 a+2 b .$ Then,
\[
\begin{aligned}
\nabla f &=\lambda \nabla g \Rightarrow\left\langle\frac{\partial A}{\partial a}, \frac{\partial A}{\partial b}, \frac{\partial A}{\partial \theta}\right\rangle=\lambda\left(\frac{\partial q}{\partial a}, \frac{\partial g}{\partial b}, \frac{\partial g}{\partial \theta}\right) \\
& \Rightarrow\langle b \sin \theta, a \sin \theta, a b \cos \theta\rangle=\lambda\langle 2,2,0\rangle \\
\Rightarrow \lambda=\frac{b}{2} \sin \theta=\frac{a}{2} \sin \theta \text { and } a b \cos \theta=0
\end{aligned}
\]
In case that $a=0,$ or $b=0,$ or $\theta=0,$ the resulting minimum area is $A=0$ Therefore, we consider the remaining possibilities
\[
b \text { and } \theta=\pi / 2=a
\]
We find: $a=b=\ell / 4$ and Area
$a^{2}=(l / 4)^{2}=\ell^{2} / 16=A$
