Help with solving word problems pertaining to limits of a function. This is my first post, and I apologise if I make any mistakes in setting this out, but I am just getting used to formatting. 
So In my current math unit, we are being assigned problems to do with maxima and minima of a function. Specifically, the one I am dealing with is a 2D shape maximisation problem.
So if i have a particular area $y$ that I need to maximise, and $x$ amount of material to enclose it, how am I meant to mathematically derive the maximum?
It couldn't be as simple as $x$ divided by the sides needed for the enclosure, squared to get the area, would it? I feel like there is a derivative operation needed, but am unsure how to do that when I haven't even got a starting formula to derive from.
Again, if there is an issue with my formatting, or I am asking an inane question, please be civil. I am only learning.

I'm going to suppose that you're working with a rectangle, hopefully you can generalise to other shapes as necessary.
For an $l \times b$ rectangle, the area is $y = lb$ and the perimeter is $x = 2(l + b)$. Since we have a fixed amount of material, $x$ is a known value, so $l$ and $b$ are related by $l = \frac{x}{2} - b$, meaning that $y = b\left(\frac{x}{2} - b\right)$ expresses $y$ as a function of $b$.
The question, then, is what is the maximum value of $y$, and what value of $b$ is it obtained at? Well, we can differentiate $y$ with respect to $b$ and set that to zero:
$$\begin{eqnarray}\frac{dy}{db} & = & \frac{d}{db}\left[b\left(\frac{x}{2} - b\right)\right] \\
& = & \frac{d}{db}\left(\frac{bx}{2} - b^2\right) \\
& = & \frac{x}{2} - 2b \end{eqnarray}$$
So we have an extreme value for the area when $\frac{x}{2} - 2b = 0$, i.e. when $b = \frac{x}{4}$. And when that happens, $y = \frac{x}{4}\times\frac{x}{4} = \frac{x^2}{16}$. Notice that $b = l = \frac{x}{4}$, meaning that the shape is a square.
You can do a little work with second derivatives, or graphs, or whatever's your preference, to show that this particular extreme value is a maximum.
