Help with a pre-calculus problem, (I'm not sure how to finish solving this problem)

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Four hundred feet of fencing is to be used. Find the dimensions of the playground and maximize the total enclosed area. What is the maximum area?

The first thing I did was use x for the width and y for the length of the playground. There are 2 sides that can be labelled as x and since there is a third part of fencing that divides the playground in 2, there are 3 sides that can be labelled as y.
So I got the formula 2x+3y = 400

I then isolated the y term:

3y = 400 - 2x

y = 133.33- 2/3x

I'm kind of stuck here, I'm not entirely sure if this is right, either. What do I do after this?

So next we need a formula for Area. In your nomenclature, area = x*y, so you have y isolated, so plug in that value and we have Area = x*(400/3 - 2x/3) = (400/3)x - (2/3)x².

This is a parabola opening down. The maximum is at the vertex. A parabola's vertex is midway between the zeros.

Solving we have x = 0 and x=200 for the zeros. So halfway between this is x=100.

You can verify this is a maximum with a plot of the parabola, or you can put it in the form (from Algebra I, which I don't remember off the top of my head) and see where the vertex is that way.

You can solve for y with y = 400/3 - (2/3)(100) = 66 2/3.