Fencing optimization problem for calculus

A farmer wants to build a rectangular corral split in 3 with an area of 900 meters squared. What are the dimensions needed to build the corral that requires the least amount of fencing?
(it's a rectangle with 2 horizontal lines cutting throuh it to make 3 pieces)

All i know is to find the derivative for the fencing equation to get a minimum.

To start out, we create an equation to determine the area of the rectangle:

A = lw

We know we want it to be 900m², so we can plug that into our area, leaving two variables:

900 = lw

Now we want to use the least amount of fencing material to do this. So we know we'll have 2 lengths and 2 widths to create the perimeter, but then we have to more horizontal lines to worry about, so we'll double the lengths and call it 4.

Ultimately, doens't matter which variable gets doubled, you'll get the same result in the end, just values different in l and w. But since you care about the sum of l and w, not the values individually, it doesn't matter what each are.

So we have:

P = 4l + 2w

Now we have two equations and three unknowns. We want to turn them into one equation with two unknows so we can do the derivative. So we'll solve for either l or w in equation 1 and then plug it into equation 2:

900 = lw
900/l = w

Now plug that into the second equation:

f = 4l + 2w
f = 4l + 2(900/l)
f = 4l + 1800/l

Let's turn that 1800/l into a negative exponent:

f = 4l + 1800l^(-1)

Now to get the minimum f to make this equation fit, find the derivitive, then set it to 0 and solve. This will be either a minimum or maximum.

df/dl = 4 - 1800l^(-2)

or

df/dl = 4 - 1800/l²

Now set that to 0 and solve for l

0 = 4 - 1800/l²
1800/l² = 4
1800 = 4l²
450 = l²
l = √450
l = 15√2

I threw out the negative answer, since we're dealing with physical lengths. So we now have one value of l, We don't really need to know the total length of fencing, but if you want to plug it in and solve for it to see what f is, you can, and that will be your minimum.