
Because there are 2 dividers, the perimeter = 2l + 4w, where l is the length of the field, and w is the width of the field (and the dividers)
2l + 4w = 2400
let x = l
2x + 4w = 2400
4w = 2400  2x
w = 600  0.5x
Now you can find the area.
A = l*w
A = x(600  0.5x)
Now you need to optimize area. A calculator is the easiest way, although if you are in Calculus you would likely have to show more work.
If you can use a graphing calculator, enter the equation in and find the maximum value on the graph.
I will demonstrate the calculus way below:
A = 600x  0.5x^2
dA/dt = 600  x
Find where dA/dt = 0
600  x = 0
x = 600
dA/dt changes signs from + to  at x = 600, so there is a relative maximum at x = 600
Because x = length, the length is 600
w = 600  0.5x
w = 600  300
w = 300
The dimensions are 300m * 600m

let the width of the whole rectangle be x m (there will be 4 of these)
let the length be y m
then 4x + 2y = 2400
2x + y = 1200
y = 1200  2x
Area = xy
= x(12002x)
= 2x^2 + 1200x
Now, I don't know if you are studying Calculus.
If you do, then
d(Area)/dx = 4x + 1200
= 0 for a max area
x = 300
then y = 600
and the max area is (300(600) = 180000
If you don't know Calculus, complete the square on the above quadratic
you should end up with
Area = 2(x300)^2 + 180000

Hey, we think alike, lol


Interesting...I did not think to complete the square to find the maximum area though. Based on steven's other questions I was not sure whether he was in Calculus.

a farmer has a rectangle field with an area of 3/4 square like the field is 1/2 mule wide what is the length of the field