Question

Prove that of all rectangles with given area, the square has the smallest perimeter.

Answer 1

Let $x,y$ be the length and breadth of rectangle whose are is $A$ and perimeter is $P$.

$\Rightarrow$ $P=2(x+y)$ --- ( 1 )

We know, $A=x\times y$

$\therefore$ $y=\frac{A}{x}$ ---- ( 2 )

Substituting value of ( 2 ) in ( 1 ) we get,

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

For maximum or minimum values of perimeter $P$

$\Rightarrow$ $\frac{dP}{dx}=2(1-\frac{A}{x^2})=0$

$\Rightarrow$ $1-\frac{A}{x^2}=0$

$\Rightarrow$ $x^2=A$

$\Rightarrow$ $x=\sqrt{A}$ [ Dimension of rectangle is always positive ]

Now, $\frac{d^{2}P}{d{x}^2}=2(0-A\times \frac{(-1)}{x^3})=\frac{2A}{x^3}$

$\therefore$ ${\left[\frac{d^{2}P}{d{x}^2}\right]}_{x=\sqrt{A}}=\frac{2a}{(\sqrt{A}{)}^3}>0$

i.e., for $x=\sqrt{A},P$ (perimeter of rectangle ) is smallest.

$y=\frac{A}{x}=\frac{A}{\sqrt{A}}=\sqrt{A}$

Hence, for smallest perimeter, length and breadth of rectangle are equal $(x=y=\sqrt{A})$ i.e., rectangle is square.