Find the maximum area of the rectangle that can be formed with fixed perimeter $20$.

Let length = x ; Breadth = y

Given perimeter = 20

$2(x+y)=20$

$x+y=10$

$y=10-x.....\left(1\right)$

area of rectangle $A=xy$

$A=x(10-x)=10x-x^2$

Let $f\left(x\right)=10x-x^2$

$f^{\mathrm{\prime }}\left(x\right)=10-2x\Rightarrow f^{\prime \prime }\left(x\right)=-2<0(max)$

Clearly it has maximum area 

Consider $f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow 10-2x=0\Rightarrow x=5$

sub in $\text{ (1) }y=10-x=10-5=5$

max area $=xy=5\left(5\right)=25\text{ sq. units }$