Answer :
the way I did it before I knew calculus, was that when legnth=width, you get max area with minimumperimiter
so L=W= √2000=20√5
legnth and width should be 20√5 meters
the following is calculus
xy=2000 and 2(x+y)=P
solve
ok so
xy=2000
divide both sides by x
y=2000/x
sub 2000/x for y in other equation
2(x+2000/x)=P
2x+4000/x=P
to find the minimum value of this, take the derivitive and find where it equals 0
2-4000/(x^2)=0
2=4000/(x^2)
2x^2=4000
x^2=2000
x=√2000
x=20√5
y=2000/x
y=2000/(√2000)
y=√2000=20√5
x=y=20√5 meters
so L=W= √2000=20√5
legnth and width should be 20√5 meters
the following is calculus
xy=2000 and 2(x+y)=P
solve
ok so
xy=2000
divide both sides by x
y=2000/x
sub 2000/x for y in other equation
2(x+2000/x)=P
2x+4000/x=P
to find the minimum value of this, take the derivitive and find where it equals 0
2-4000/(x^2)=0
2=4000/(x^2)
2x^2=4000
x^2=2000
x=√2000
x=20√5
y=2000/x
y=2000/(√2000)
y=√2000=20√5
x=y=20√5 meters