If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me.
This mentality is what's probably bothering people.

But, in your defense, you probably have an auditory or kinesthetic learning type so you need a different approach to really learn this stuff. It's basically what leads to the feeling that you know a little bit of what you need to do as far as writing it down but then you stare at it and don't really know what you're doing with it, like here:

I know the volume of an open box is (l-2x)(w-2x)(x). I just don't know what to do with the 1200 cm^2.
Let's start you off on the right path. Let's say y = 3 - 4x - x^2 and we want to maximize y. Well how do we find maximums and minimums with Calculus? Derivatives. So y' = -4 - 2x, let y'=0 to find the maximum x value, so 0 = -4 - 2x and x = -2. Then our maximum y is 3 - 4(-2) - (-2)^2 = 7. Graph y = 3 - 4x - x^2 to verify and maybe that will help your understanding as well.

Edit: Waiting on my girl to get out of the shower so graphed it for you...



Note that the maximum y-value is 7, just like we figured out. (/edit)

Here's another for you to think about: Suppose we're going to fence in a rectangular area with 100 feet of fencing but we're fencing beside of a river so we only need to do three sides, and we want to maximize the area we can fence in. Here's a picture to make sure you get the idea:



Well we need an equation for area first and foremost. The area of a rectangle is length times width, so let's define those. If we let the left side of the fence be length x, then the opposite side is also length x, and the top side is 100 - 2x (since we have 100 feet total of fencing). Now we have an equation for area: A = width * length = x(100-2x).

We want to maximize area here, so how would we do that?