Step 1:

Let the given area be A and x, $\large\frac{A}{x}$ the sides of a, rectangle (Since area= length x breadth, the length and breadth are $x, \large\frac{A}{x}$)

Step 2:

The perimeter of the rectangle =2 (length+breadth)

$p=2 (x +\frac{A}{x})$

Step 3:

At extreme values of $p , \large\frac{dp}{dx}$$=0$

$\large\frac{dp}{dx}$$=2(1-\large\frac{A}{x^2})$

$\therefore 2(1-\large\frac{A}{x^2})$$=0$$=> x= A^{1/2}$

Now $\large\frac{d^2p}{dx^2}$$= 2 \times \large\frac{2A}{x^3}=\large\frac{4A}{x^3}$

$\large\frac{d^2p}{dx^2(x=A^{1/2})}$$= \large\frac{4A}{A^{3/2}}=\large\frac{4}{\sqrt A}$$ > 0$

$\therefore x=A^{1/2}$ corresponds to a minimum value of P

Step 4:

The sides of the rectangle of minimum perimeter are $ A^{1/2},A^{1/2}$ =>it is the square