Finding the conditional extrema | Correct Method Find the conditional extrema of $ f(x,y) = x^2 + y $ subject to $x^2 + y^2 = 4$
Do you I use the Lagrange multipliers for this?
Or do I use the partial differential method and figure out what side of the zero-point this equation falls under:
$$ f_{xx}f_{yy} - {f_{xy}}^2 $$ ?
 A: We have:
$$\tag 1 f(x,y) = x^2 + y ~~ \mbox{subject to} ~~\phi(x) = x^2 + y^2 = 4$$
A 3D plot shows:

If we draw a contour plot of the two functions, we get:

From this plot, we see four points of interest, so we will use Lagrange Multipliers to find those and then we just need classify them as local or global min or max or not classifiable.
We can write:
$$\tag 2 F(x,y) = f + \lambda \phi = x^2 + y + \lambda(x^2 + y^2)$$
So, 
$\tag 3 \dfrac{\partial F}{\partial x} = 2x (1 + \lambda) = 0 \rightarrow x = 0~ \mbox{or}~ \lambda = -1, ~\mbox{and}~$
$\tag 4 \dfrac{\partial F}{\partial y} = 1 + 2 \lambda y = 0 $
From $(3)$, we get:
$$x = 0 \rightarrow y^2 = 4 \rightarrow y = \pm 2 \rightarrow \lambda = \pm \dfrac{1}{4}$$
From $(3)$ and $(4)$, we get:
$$\lambda = -1, y = \dfrac{1}{2} \rightarrow x^2 + \dfrac{1}{4} = 4 \rightarrow x = \pm \dfrac{\sqrt{15}}{4}$$
Summarizing these results, we have the four potential critical points to investigate at:
$$(x,y) = (0,2), (0,-2), \left(\dfrac{\sqrt{15}}{2},\dfrac{1}{2}\right),\left(-\dfrac{\sqrt{15}}{2},\dfrac{1}{2}\right)$$
Now, classify these critical points using the typical method.
You should end up with (you can actually make these out from the 3D plot above):


*

*Local min at $(0,2)$

*Global min at $(0,-2)$

*Global max at $\left(\pm \dfrac{\sqrt{15}}{2}, \dfrac{1}{2}\right)$

