2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.