<s>[INST] Optimization Triangle I can't seem to finish this task. I found the derivative and the critical numbers, except when the derivative $= 0$. I don't understand how I can find it manually.
Task:

$2$ points A and C lie on a line $l$. a point B is $4$ away from line $l$. The sum of AB + BC is always equal to $10$. Find the shortest distance for AC. (A and C can not be on the same side of the perpendicular, however, A or C can be on the perpendicular).

What I got:
I have these equations
let AB $= x$, and BC $= y$.
$$\text{AC} = \sqrt{x^2 - 4^2} + \sqrt{y^2 - 4^2}$$
$$\text{AC} = \sqrt{x^2 - 16} + \sqrt{y^2 - 16}$$
$$\text{AC} = \sqrt{x^2 - 16} + \sqrt{x^2 - 20x + 84}$$
$$\text{AC}' = \frac{x}{\sqrt{x^2 - 16}} + \frac{(x - 10)}{\sqrt{x^2 - 20x + 84}}$$
$$\text{AC}' = \frac{x}{\sqrt{(x+4)(x-4)}} + \frac{x - 10}{\sqrt{(14-x)(6-x)}}$$
I can now find that $x = 4, x = -4, x = 14, x = 6$ are critical numbers, and only $x = 4$ and $x = 6$ apply. But how do I now find AC' $= 0?$
 [/INST] From $$0 = \frac{x}{\sqrt{(x+4)(x-4)}} + \frac{x-10}{\sqrt{(14-x)(6-x)}},$$ simply multiply both sides by $\sqrt{(x+4)(x-4)} \cdot \sqrt{(14-x)(6-x)}$ to get
$$0 = x\sqrt{(14-x)(6-x)} + (x-10){\sqrt{(x+4)(x-4)}}.$$
Move one of the two terms to the left hand side, and then square both sides to get
$$x^2(14-x)(6-x) = (x-10)^2(x+4)(x-4).$$
Expanding, we get
$$x^4 - 20x^3 + 84x^2 = x^4 - 20x^3 + 84x^2 + 320x - 1600.$$
This boils down to $x = 5$.
The most tedious part is the expansion step, but it is still doable by hand.
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