Okay so I have a quadratic function problem. I will omit the problem for now just because we don't really need it. My problem is: M is surface area. Do I have to write M(x, y) or just M in the area I've bolded? Here is my work:

Problem 2

x = Length of a side of the base (ft)

y = Height of rectangular box (ft)

M = Surface area of rectangular box (ft2)

This question asks for the largest surface area possible with a rectangular box that has a square base and the sum of whose 12 edges is 8 ft. To determine that, we must first find the formulas with the given information for the sum of the edges and the surface area. The area of a square with side length x is x2, and the area of a rectangle with side lengths x and y is xy. There are four edges of the rectangle, and eight edges of the base.

Sum of the edges:

4y + 8x = 8

4y = 8 – 8x

y = 2 – 2x

Surface area:

M = 2x2 + 4xy

M = 2x2 + 4x(2-2x)

M = -6x2 + 8x

And that is the quadratic function for the surface area.

We now must determine the value for x, or the side length of the base, that will produce the largest surface area. To do this, we will find the x value of the vertex, determined by x = -b/2a.

a = -6, b = 8

x = -8/-12 = 2/3

The side length of the base that will yield the maximum surface area is 2/3 ft. Now, to find y, or the height of the box, we must plug in x to the equation y = 2 – 2x.

y = 2 – 2(2/3) = 2/3

The height of the box that will yield the maximum surface area is 2/3 ft.

The box must be a square with side lengths 2/3 ft to have the maximum surface area.

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Technically, $M$ is a function of $x$ and of $y$: $M(x, y)$. But using $M$ is just find for calculations, because, it becomes, after substituting, a function of $x$. So I see no problems with using only "$M$" to represent surface area.
The ONLY thing I'd change is one word(s) in your conclusion, and adding the actual maximum surface area: "The box must be a [cube] with side lengths $\large\frac 23$ ft. to have the maximum surface area, [which is $<$insert calculation of M$>$ ft$^2$.]