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! 

the constraint is $x + y = 12.$  at a local extremum of $\sqrt{x^2 + 4} + \sqrt{y^2 + 9},$  the critical numbers satisfy $$dx + dy =0,\quad \frac{x\, dx}{\sqrt{x^2 + 4}} + \frac{y\, dy}{\sqrt{y^2 +9}} = 0 \to x\sqrt{y^2 + 9}=y\sqrt{x^2 + 4} $$  squaring the last equation we have $$9x^2 =4y^2 \to y = \pm\frac32x, x + y = 12 \implies x=24/5, y = 36/5,\\ \sqrt{x^2 + 4} + \sqrt{y^2+9} = 13$$ therefore the global minimum of $$\sqrt{x^2 + 4} + \sqrt{y^2+9} \text{ is } 13.$$
