In this equation, if two of the variables are held constant, which variable will bring out the maximum positive change in the answer? I tried doing this in excel, but I'm having trouble figuring out if I'm on the right track. I got 27 different scenarios in the excel sheet and sorted by the highest value. Instinct tells me this is not the correct way to determine this.
$$S = \frac{AB}{1+C-B}$$
I've renamed your variables so as to be more familiar looking: $$f(x,y,z) = \frac{xy}{1+z-y}$$ In multivariable calculus, there's something called the gradient of a function $f$ in a point $a$, which is $$\nabla f(a) = \left. \pmatrix{\partial_xf\\\partial_yf\\\partial_zf} \right|_a$$ which means that the 1st/2nd/3rd entry in the vector is the derivative of $f$ with respect to $x$/$y$/$z$, respectively, evaluated in the point $a$.
All right, how do we use it? Well, there's a theorem that states that when you take $f$ in some point $a$, the direction in which $f$ grows the most from there is the direction of the gradient in that same point.
In your case, the gradient is $$\nabla f(x,y,z)=\pmatrix{\frac{y}{1+z-y} \\ \frac{x(1+z)}{(1+z-y)^2} \\\frac{-xy}{(1+z-y)^2}} = \frac{x}{(1+z-y)^2}\pmatrix{\frac{y}{x}(1+z-y) \\ 1+z \\-y}$$
So if you take any point $(x,y,z)$ and plug it into that gradient, whichever of the three is bigger is the one that will bring the maximum change. For instance, if it was the first that was the biggest, then you know that you should be going in the $x$-direction to get the biggest change.
Try playing around with it and - if you have the possibility - make some contour plots... it's fun, I promise!
EDIT: Visualization time!
This would be so easy if only had two variables. With three variables, however, we need three axis ($x,y,z$) and one axis for the value of $f(x,y,z)$ - that's one too many!
Still, there are a couple of methods that can aid with the visualization. The one I'll choose here is one where you only look at one value of $f$ at a time. If we call this value $c$, we can write $$f(x,y,z)=c=\frac{xy}{1+z-y} \leftrightarrow xy = c(1+z-y) \leftrightarrow z=\frac{xy}{c}+y-1$$ Now we have a function $z(x,y)$ of only two variables, and this can be plotted, here for $c=1$ (note that $c$ doesn't change the shape of the surface, only how steep it is):
So how to interpret this? In relation to $f$ this is, admittedly, a bit hard. What we can say is that, for a constant value of $f$, $x$ and $y$ influences $z$ equally.
Let's try something more powerful: Plotting the gradient! Since the gradient in a point is a vector (an arrow, with a magnitude and a direction), its generalization to any point is a vector field. I've plotted it here:
The $x,y$-axis lies closest to the screen ($y$ is up, $x$ is to the left (notice the axis-values!)) and the $z$-axis is going into the screen.
The way to read this is as follows: Pick a point $(x,y,z)$ and look at the arrow coming out of this point. The direction the arrow is pointed is the direction you'd want to take a step in to make the value of $f$ bigger. The magnitude of the arrow indicates how big a change a single step results in (so big arrow $\rightarrow$ big effect, and vice versa). Neat, right?
Since the directions can be hard to see, I've the same thing, but seen from the left instead of from the right:
Note that I've chosen to plot in the interval $[0,5]$ for all three variables. However, the behavior of the vector field seems to scale without changing when looking at larger intervals, so I think it is fair to assume the analysis below is valid for larger values as well.
As you can see, the gradients all seem to point in pretty much the same direction, namely in the direction of lower $z$ and higher $y$. $x$ only seem to vary the magnitude of the gradient, i.e., how big an effect taking a step has.
Let's take a look at a specific point, say $(4,4,4)$. Then $$\nabla f(4,4,4) = 4\pmatrix{1 \\ 5 \\ -4} \rightarrow r_{4,4,4} = \frac{1}{\sqrt{42}}\pmatrix{1 \\ 5 \\ -4}$$
where $r_{x,y,z}$ is just the unit vector pointing in the direction of $\nabla f(x,y,z)$.
Let's compare this with the gradient of another point $(4,3,2)$. Then $$r_{4,3,2} =\frac{1}{\sqrt{2}}\pmatrix{0 \\ 1 \\ 1}$$
It is clear that these do not point in the same direction. From this can be concluded that what direction you should go in to make $f$ bigger depends on where you're standing. But in general, from our purely visual analysis regarding the general direction of the field of the gradient, we can say:
In general, if you want to maximize $f$, go in the direction of higher $y$'s and lower $z$'s. To get a bigger overall effect of your steps, increase $x$.
And to make the conclusion complete:
But really, which way you should go depends on the specific values of $(x,y,z)$.