Let length of adjacent sides of rectangle is $x cm$ and $y cm$ so perimeter of rectangle is $p=2(x+ y) cm$ and the area $A=x y cm ^{2}$
$\Rightarrow A=x\left(\frac{p}{2}-x\right)$
for maxima
$\frac{d A}{d x}=0 \Rightarrow \frac{p}{2}-2 x=0$
$\Rightarrow x=\frac{p}{4} cm$
and $y=\frac{p}{4} cm$
$\therefore $ For given perimeter of rectangle $p$, the maximum possible area
$A=\frac{p}{4} \times \frac{p}{4}=\frac{p^{2}}{16} cm ^{2}$