Why is alternative approach to constrained optimization incorrect? I am studying for the math GRE subject test, and my practice exam has a problem that goes as follows:
Find the minimum distance from the origin to the curve $3x^2 + 4xy + 3y^2=20$.
Apparently I was supposed to solve this using Largrange multipliers, and I will make sure to try that as well. But I didn't do that because when I looked at the problem I saw what I thought would be a simpler method of solving it. I noticed that
$3x^2 + 4xy + 3y^2 = 20 = x^2+y^2 + 2(y+x)^2 = D^2 + 2(y+x)^2$
Therefore, in order to minimize the distance, which is equivalent to minimiziong $D^2$, we can simply minimize $f(x,y) = 20 - 2(y+x)^2$. When I take $f_x=0$ and $f_y=0$, I get the same 2 equations: $-4(x+y) = 0$, which implies that $x=-y$.
However, it turns out that after plugging this in, this is not the answer. In fact, when you do it with lagrange multipliers, you get that not only is $x=-y$ is a critical point, but also $x=y$. It turns out that x=y is the one that actually yields the minimum.
So my question is, where did I go wrong? Why does my solution method leave out one of the critical points?
Any help is greatly appreciated.
Best,
Paul
Edit: I have found the problem with my method, and as Nick points out below it is that there are in fact two different ways to complete the square. I have a feeling that something slightly deep is going on here, and if anyone could illuminate further I would really appreciate it. 
 A: I apologize if this answer is not detailed enough as I do not have a thorough solution to your problem myself. But I do believe that you leave out one critical point because the way you create the square is somewhat arbitrary. For example, you could have written 
$3x^2 + 4xy + 3y^2 = 20 = 5(x^2 + y^2) - 2(x-y)^2=5D^2 - 2(x-y)^2$
so then you need to minimize $f(x,y) = 20 + 2(x-y)^2$, which gives you $x=y$. That is your second critical point.
A: If $(x, y)$ is unconditional optimum then it is necessary the solution to $\nabla f = 0$, but in your case when minimizing $f(x, y)$ you still have constrainted problem and $\nabla f = 0$ is not the necessary condition for the optimum in such kind of problem. Also note that the solution $x = -y$  of $\nabla f = 0$ is the global maximum for $f(x, y)$. 
So you go wrong when trying to solve 
$$f(x, y) \to \min, \text{ subject to } 3x^2 + 4xy + 3y^2 - 20 = 0,$$ as unconstrainted optimization problem 
$$f(x, y) \to \min,\ (x,y) \in \mathbb{R}^2.$$
A: There is a way to do what you wanted, but you made a bit of a problem for yourself by imbedding the "distance-squared-from-the-origin" function into the constraint curve equation.  Instead, the problem would be dealt with by using the constraint to "simplify" the function expression.  So we might use
$$ 3x^2 \ + \ 4xy \ + \ 3y^2 \ = \ 20 \ \ \rightarrow \ \ y^2 \ = \ \frac{20}{3} \ - \ x^2 \ - \ \frac43 xy $$ $$ \Rightarrow \ \ D^2 \ = \ x^2 \ + \ y^2 \ = \ \frac{20}{3} \   - \ \frac43 xy \ \ . $$
As the other posters have said, if one were to use the Lagrange method, we would still need to use this together with the constraint equation, since this result alone (as with the equation you developed) is only one equation in two variables.  With the modified function here, the Lagrange equations are
$$ 6x  \ + \ 4y \ = \ \lambda·\left(-\frac43y\right) \ \ , \ \ 4x  \ + \ 6y \ = \ \lambda·\left(-\frac43x\right)  $$ $$ \lambda \ = \ -\frac{9x \ + \ 6y}{2y} \ \ = \ \ -\frac{6x \ + \ 9y}{2x} \ \ \Rightarrow \ \ 18x^2 \ + \ 12xy \ \ = \ \ 12xy \ + \ 18y^2 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ \ , $$
with the constraint equation then leading to minimum and maximum distances for points on the curve.
But for a timed test like the GRE (which I hope went well for you, since this is an old question), there is something quicker that can be done.  Since the distance-squared function and the constraint curve equation have symmetry about the origin, the points at which extremal values for the function occur are symmetrically located on the lines $  \ y = x \ $ and $ \ y = -x \ \ . $  The constraint equation, which described a "rotated ellipse" centered on the origin, will then give two pairs of points:
$ \ \mathbf{ y \ = \ x } \ :   \quad 3x^2 \ + \ 4·x·x \ + \ 3x^2 \ = \  20 \ \ \Rightarrow \ \ 10x^2 \ = \ 20 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ = \ 2 \ \ ; $
$ \ \mathbf{ y \ = \ -x } \ :   \quad 3x^2 \ + \ 4·x·(-x) \ + \ 3(-x)^2 \ = \  20 \ \ \Rightarrow \ \ 2x^2 \ = \ 20 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ = \ 10 \ \ . $
Since all the GRE problem asks for is the minimum distance from the origin for a point on the ellipse, we have $ \ D_{min} \ = \ \sqrt{2+2} \ = \ 2 \ \ . $  (We also have the maximal distance $ \ D_{max} \ = \ \sqrt{10+10} \ = \ 2\sqrt5 \ \   $ and can locate the corresponding points, but those results weren't requested.)

