Let length of farm = $x$ meter
Width of farm = $y$ meter
Area = $xy$ =100
Parameter = $P$ = $2(x+y)$ = $2(x+\frac{100}{x})$
$\frac{\mathrm{d} p}{\mathrm{d} x}$ = $2(1-\frac{100}{x^2})$ =$0$
So $x$ = 10 Meter // using Maxima/minima concept we can see that's point of minima for perimeter
So minimum Perimeter =$2(10+10)$=$40$
Graph would be kind of Hyperbolic in +ve xy axis so continuous mean every parameter values that is greater than or equal to 40 is possible.