Finding lowest value measures between three locations 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.
$$\left(\frac{7K}{x^2 + 10}\right)' = -\frac{14K}{(x^2 + 10)^2}$$

Choose a co-ordinate system with Plant A as origin.
Basic trigonometry to find the angles of the triangle ABC
You will get (in Radians)
$\angle A = 0.9582$
$\angle B = 0.4214$
$\angle C = 1.762$  
Now the position of Plant C is straightforward $(5\cos A, -5\sin A)$
Assuming the origin is fixed at Plant A the amount of pollution at any arbitary point (x, y) is 
$\frac{K}{x^2 + y^2 + 10}+ \frac{K}{4((x-12)^2 + y^2 + 10))} + \frac{K}{2((x-2.874968)^2+(y+4.0907895)^2+10))}$
On Road AB this reduces to $\frac{K}{x^2+10}+ \frac{K}{4((x-12)^2+10))} + \frac{K}{2((x-2.874968)^2+26.7345587))}$
Factoring out K and minimizing $\frac{1}{x^2+10}+ \frac{1}{4((x-12)^2+10))} + \frac{1}{2((x-2.874968)^2+26.7345587))}$
Minimum occurs at $x \approx 8.7758$ miles from Plant A on Road AB
