Question

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

Answer 1

let the sides of rectangle be $l,b$ length and breadth respectively.let the given area be $A$.

$A=l\times b$

$b=\frac{A}{l}$

$perimeter=2\times (l+b)$

$perimeter=2\times (l+(\frac{A}{l}))$

let the function $f\left(l\right)$ be the perimeter.smallest value of $f\left(l\right)$

is;

$f\left(l\right)=2\times (l+(\frac{A}{l}))$

$\frac{d}{dx}f\left(l\right)=2\times (1-(\frac{A}{l^2}))$

for a minimum value of a function its differential should be zero;

$\frac{d}{dx}f\left(l\right)=0$

$2\times (1-(\frac{A}{l^2}))=0$

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

$1=\frac{A}{l^2}$

$l^2=A$

$b=\frac{l^2}{l}$

$b=l$

we know that when the length,breadth equals it becomes a square.