Question 1109885
I believe what is meant (but not really clearly specified) is shown in the sketch below:
 
{{{drawing(340,160,-20,320,-20,140,
rectangle(0,0,300,120),rectangle(100,0,200,120),
locate(3,65,W),locate(103,65,W),
locate(203,65,W),locate(303,65,W),
locate(114,-2,L=540-2W),arrow(110,-10,0,-10),
arrow(190,-10,300,-10)
)}}}
 
Numbers/variables below are meant to be lengths in feet and areas in square feet,
but I am not writing the units repeatedly.
{{{fencing}}}{{{length}}}{{{"="}}}{{{2L+4W}}}{{{"="}}}{{{1080}}} ,
so {{{2L=1080-4W}}} --> {{{L=(1080-4W)/2}}} --> {{{L=540-2W}}} .
The total surface area is
{{{36000=(540-2W)*W}}}
{{{36000=540W-2W^2}}}
{{{2W^2-540W+36000=0}}}
Dividing everything by 2, we get the simplified, equivalent equation
{{{W^2-270W+18000=0}}}
Factoring to solve that equation, we get
{{{(W-150)(W-120)=0}}} --> {{{system(W=120,"or",W=150)}}} .
So, either {{{system(W=120,L=36000/120)}}} --> {{{system(W=120,L=300)}}} , or
{{{system(W=150,L=36000/150)}}} --> {{{system(W=150,L=240)}}}
That gives us two possible answers for the dimensions of "the whole enclosed rectangle."
Either the whole enclosed area is {{{120ft}}} wide by {{{300ft}}} long, or  it is
{{{150ft}}} wide by {{{240ft}}} long.