# Is this argument that $(0,0)$ is a local minimum for $f(x,y)=x^4+3xy^2+y^2$ correct?

I'm trying to find a local minimum for $f(x,y) = x^4 + 3xy^2 + y^2$.

After finding that a critical point is at $f(x,y)$ is $(0,0)$, I am wondering if this is proper formal justification to prove that this is indeed the case?

Let $\epsilon >0$ be given. Let $|y|=-x$
(Is this proper? I just want to keep x and y to have the same absolute value basically)
Let $x = \max\{\frac{-1}{3}, \frac{-\epsilon}{2}\}$.

Then $\sqrt{x^2+y^2}=\sqrt{2\frac{\epsilon}{2}^2} = \frac{\epsilon}{2} < \epsilon$
(Is this proper to show that I can make x, y as close to $\epsilon$ as I want?)

Since $\frac{-1}{3}\leq x\lt 0$ then $3x \geq -1$ and $x^4+3xy^2+y^2 \geq 0$
$\therefore (0,0)$ is a local minimum of $f(x,y)$

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Please make the body of your message self-contained; don't realy on the title for content. Think of the title as the spine of a book: useful to find it later, but not really part of the narrative. – Arturo Magidin Mar 10 '11 at 3:18
Ok I'll keep that in mind! – Sam C Mar 10 '11 at 3:19
In any case, +1 for showing your work! – Arturo Magidin Mar 10 '11 at 3:44

In order to show that $f(x,y)$ has a local minimum at $(0,0)$, you need to show that there is an $\epsilon\gt 0$ such that for all $(x,y)$, if $\sqrt{(x-0)^2+(y-0)^2}\lt\epsilon$ (that is, if $(x,y)$ is $\epsilon$-close to $(0,0)$), then $f(0,0)\leq f(x,y)$. In other words: there is an open ball around $(0,0)$ with the property that $f(0,0)$ is the smallest value that $f$ takes in that ball.

So you are not "given an $\epsilon\gt 0$", as you are with limits. Instead, you get to pick the $\epsilon$.

However, you don't get to restrict the values of $x$ and $y$: any value of $x$ and $y$ that falls within $\epsilon$ of $(0,0)$ has to be accounted for. So you don't get to say "Let $|y|=-x$." And you don't get to pick $x$, either.

So, I'm afraid that this is not a valid way to go about it.

Instead, let's think about $f(x,y)$. Both the $x^4$ and the $y^2$ will contribute positive factors; so really we only have a potential problem (i.e., a value of $f$ smaller than $0$) when $x\lt 0$; if $y=0$ the factor $3xy^2$, the only possible source of problems, vanishes, so we just need to worry about the case $y\neq 0$. Then we have $f(x,y) = x^4 + 3xy^2 + y^2 = x^4 + y^2(3x+1)$.

This means that if $3x+1\geq 0$, then $f(x,y)\geq 0 = f(0,0)$, even if $x$ is negative and has absolute value much larger than $y$. And this will be achieved if $x\geq - \frac{1}{3}$. So all you need to do is make sure that your $x$ is larger than $-\frac{1}{3}$ (as you probably noticed, given the value you wanted to assign to $x$).

So: all you need to do is find some $\epsilon$ with the property that if $\sqrt{x^2+y^2}\lt \epsilon$, then you'll know that $x\geq -\frac{1}{3}$. Then you can argue that for that $\epsilon$, you have the property you want (every point $(x,y)$ within $\epsilon$ of $(0,0)$ will have $f(x,y)\geq f(0,0)$). How do you find the $\epsilon$? Draw a picture: $\epsilon$ is the radius of a circle around $(0,0)$; you want every point inside the circle to have $x$-coordinate greater than or equal to $-\frac{1}{3}$. What value(s) can you pick for $\epsilon$? (There is no single answer, since if you have a value that works, any smaller value will also work).

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@Sam: No, the point is a local minimum, you just went about showing it is the wrong way. And no, you cannot pick $x$ to make it work the way you think you can make it work. See my expanded answer. – Arturo Magidin Mar 10 '11 at 3:32
Thank you very much sir. I will up vote when I get enough reputation. – Sam C Mar 10 '11 at 3:56