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.



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.


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.

  • $\begingroup$ Thank you, this is exactly the type of answer I was looking for. The method of completing the square is perhaps arbitrary but not completely arbitrary -- as far as I can tell there are only two ways of doing it, and these two ways give the two critical points. $\endgroup$ – Paul Sep 1 '14 at 23:18
  • $\begingroup$ You are welcome. Also, even though you have probably figured it out too, I just noticed that the function you were trying to minimize actually has no minimum, you found the maximum since it is concave. $\endgroup$ – Nick Sep 1 '14 at 23:20
  • $\begingroup$ yes I did find that out and also Random Jack pointed that out. I suppose my question should have been more specifically "why am I not getting both critical points?", rather than "why is my solution incorrect?" $\endgroup$ – Paul Sep 1 '14 at 23:22

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 $$f(x, y) \to \min,\ (x,y) \in \mathbb{R}^2.$$

  • $\begingroup$ I follow what you are saying... in a constrained problem the global max/min does not have to have grad(f)=0. On the other hand I am still confused. If the distance is truly equal to the expression on the right hand side, how is it still a constrained optimization problem? Doesn't it then become just a regular old optimization problem? $\endgroup$ – Paul Sep 1 '14 at 22:35
  • $\begingroup$ @Paul: It is a constrainted problem since you want to minimize the distance or as you noted equivalently minimize $f(x, y)$ on the set of points of your curve, that is subject to $3x^2 +4xy + 3y^2 - 20 = 0$, which is the constraint. When you have found equivalent function to minimize, the set of constraints have not changed, so you need to minimize $f$ on the curve but not on the whole plane and still have a constrainted problem. $\endgroup$ – Random Jack Sep 2 '14 at 3:11
  • $\begingroup$ In other words, you have found equal and more convinient representation $f(x,y)$ of the distance to zero, when points belong to your curve, but minimization is still on this curve. You have used the constraints in order to deduce that distance equals $f$, but that does not mean you can forget about the constraints after that because your transformation of distance equals $f$ only in case if $(x, y)$ belongs to the curve, hence minimization is also only on the curve. Hope this helps! $\endgroup$ – Random Jack Sep 2 '14 at 3:28
  • $\begingroup$ but doesn't (x,y) always belong to the curve, since all (x,y) satisfy the original constraint equation? $\endgroup$ – Paul Sep 2 '14 at 4:43
  • $\begingroup$ @Paul: What do you mean by "all $(x, y)$"? Only the points of the curve satisfy this equation. The main point is that $D^2 = f(x, y)$ only under the assumption that $(x, y)$ is a point of the curve. It means that you can minimize $f(x, y)$ instead of $D^2$, but with constraint "$(x, y)$ is a point of the curve". So you have the same constraints as in the initial problem. $\endgroup$ – Random Jack Sep 2 '14 at 17:27

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