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! 
 A: Just another way:
$$\sqrt{x^2+4}+\sqrt{y^2+9} \ge \sqrt{(x+y)^2+(2+3)^2} = 13$$
where we have used the triangle inequality.
A: You probably know about setting the derivative equal to $0$. The complexity of the equation we get may be discouraging. Your derivative is
$$\frac{x}{\sqrt{x^2+4}}-\frac{12-x}{\sqrt{x^2-24x+153}}.$$
I would rather write it as
$$\frac{x}{\sqrt{x^2+4}}-\frac{y}{\sqrt{y^2+9}}.$$
Nicer! Set this equal to $0$. So we get
$$\frac{x}{\sqrt{x^2+4}}=\frac{y}{\sqrt{y^2+9}}.$$
Square both sides and simplify a bit (cross-multiply). We get
$$x^2(y^2+9)=y^2(x^2+4).$$
There is some nice cancellation, and we get $9x^2=4y^2$. The rest should not be difficult.
A: Setting $f'(x)=0$ gives 
$$\frac{x}{\sqrt{x+4}}+\frac{x-12}{\sqrt{x^2-24x+153}}=0 $$
or
$$\frac{x}{\sqrt{x+4}}=-\frac{x-12}{\sqrt{x^2-24x+153}}\tag1$$
Squaring both sides of $(1)$ gives
$$\left(\frac{x}{\sqrt{x^2+4}}\right)^2=\left(\frac{x-12}{\sqrt{x^2-24x+153}}\right)^2$$
which can be rewritten as
$$\frac{x^2}{x^2+4}=\frac{(x-12)^2}{x^2-24x+153} \tag2$$
Thus, writing the numerator of the left-hand side as $x^2=x^2+4-4$ and writing the numerator of the right-hand side as $(x-12)^2=(x-12)^2+9-9$ simplifies to $(2)$
$$\frac{4}{x^2+4}=\frac{9}{x^2-24x+153} \tag3$$
Can you finish from here?
A: $$\begin{align} f'(x)=0 &\iff \frac{x}{\sqrt{x^2+4}} = \frac{12-x}{\sqrt{x^2-24x+153}}\\
&\implies \frac{x^2}{{x^2+4}} = \frac{x^2 - 24x + 144}{{x^2-24x+153}}\\
&\iff 1 - \frac{4}{{x^2+4}} = 1 - \frac{9}{{x^2-24x+153}}\\
&\iff 4(x^2 -24x + 153) = 9(x^2+4) \end{align}$$
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).
A: If you are aware of the rarely seen in action "Minkowski" inequality..then here you go: $\sqrt{x^2+2^2}+\sqrt{y^2+3^2} \geq \sqrt{(x+y)^2+(2+3)^2}=\sqrt{12^2+5^2} = \sqrt{169}=13$, and rest assured that this extrema is achievable.
A: 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 $$\sqrt{x^2 + 4} + \sqrt{y^2+9} \text{ is } 13.$$
