If a square and a rectangle have the same perimeter and their areas are $S$ and $R$ respectively, then

Let the perimeter of both the square and the rectangle be $4a$.
For the square:
Side $=a$, so Area $S=a^2$.

For the rectangle:
Let the sides be $a+x$ and $a-x$ (where $x>0$), so that the perimeter is $2[(a+x)+(a-x)]=4a$.
Area $R=(a+x)(a-x)=a^2-x^2$.

Comparing the two areas:

$$S-R=a^2-(a^2-x^2)=x^2>0$$

Therefore, $S>R$. Among all quadrilaterals with a given perimeter, the square has the maximum area.