I've made an image of what the geography of my problem looks like:
Essentially, there are two "chemical plants" (A and B) located on a road $12$ miles apart. The pollution from plant A is given by the following equation (for some constant $K$):
$\frac{K}{x^2 + 10}$
The pollution from plant B (at $x$ miles from plant B) is $1/4$ that of A.
There's also a third plant, C, which is located on the perpendicular road (the one branching off of Road A-B on the map). Plant C is $5$ miles from A and $10$ miles from B. The pollution from plant C is twice that of B.
I am trying to find the point on Road A-B where the pollution count from the three plants is minimal. I'm not really sure where to even begin with this.
The pollution from Plant B must be
$\frac{K}{4x^2 + 40},$
and from Plant C must be
$\frac{K}{2x^2 + 20}.$
Adding the three equations would yield the following:
$\frac{K}{x^2 + 10} + \frac{K}{4x^2 + 40} + \frac{K}{2x^2 + 20} = \frac{K + 4K + 2K}{x^2 + 10} = \frac{7K}{x^2 + 10}$
Finding the derivative of this would be? Not sure I'm doing this right.
$\left(\frac{7K}{x^2 + 10}\right)' = -\frac{14K}{(x^2 + 10)^2}$