find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this 
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help! 

$$\begin{align} f'(x)=0 &\iff \frac{x}{\sqrt{x^2+4}} = \frac{12-x}{\sqrt{x^2-24x+153}}\\
&\implies \frac{x^2}{{x^2+4}} = \frac{x^2 - 24x + 144}{{x^2-24x+153}}\\
&\iff 1 - \frac{4}{{x^2+4}} = 1 - \frac{9}{{x^2-24x+153}}\\
&\iff 4(x^2 -24x + 153) = 9(x^2+4) \end{align}$$
from which you find the roots (if I'm not mistaken, only one of them is valid because of the squaring in the second line).
