Question 894730
l is a bad choice for length.  Choose L for length.


A is area and is constant A=4.

{{{wL=A}}};  perimeter p is {{{p=2w+2L}}}.
{{{w=A/L}}} allows for {{{p=2(A/L)+2L}}}.


Domain for p is {{{L>0}}}.


Knowing that A=4, {{{p=2(4/L)+2L}}}
{{{p=8/L+2L}}}


Further,
{{{p=8/L+2L^2/L}}}
{{{p=(8+2L^2)/L}}}
{{{p=2(4+L^2)/L}}}


This is a rational function, and there is an asymptote for L=0.

L^2 becomes smaller faster than L, as L approaches 0, but looking at the separate terms 8/L and 2L^2/L, seeing 8/L increases while 2L^2/L will decrease without bound as L approaches 0.  The function p of L will increase without bound as L approaches 0.


{{{graph(300,300,-1,12,-1,12,8/x+2x^2/x)}}}


You might try derivative and look for the local minimum, but the graph shows a minimum perimeter at about L=2.


{{{graph(300,300,-1,4,-1,12,8/x+2x^2/x)}}}