Proving a multi variable function has an absolute maximum and minimum in an ellipse.

Question

I want to prove the function $f(x,y)=xy$ has an absolute maximum and an absolute minimum in the ellipse $E=3x^2+3y^2-2xy+4x+4y=4$ (I have already proved $E$ is an ellipse using orthogonal diagonalization).

The only critical point of $f(x,y)$ is $(0,0)$. However $x=0$ and $y=0$ does not satisfy the equation of $E$ since $-4\neq0$. Doesn't this mean that the critical point of $f(x,y)$ isn't in $E$ and therefore the function $f(x,y)$ does not have an absolute minimum or maximum in $E$? Would it be different if $E=3x^2+3y^2-2xy+4x+4y\leq{4}$?

Thank you.

Answer 1

Are you familiar with the Extreme Value Theorem for functions from $\Bbb{R}^n\to\Bbb{R}$? The ellipse $E$ is a compact set in $\Bbb{R}^2$. Therefore, the continuous function $f$ must attain an absolute max and an absolute min on that set. (This is really just the theorem that the continuous image of a compact, connected set is compact and connected, together with knowing that compact, connected sets in $\Bbb{R}$ are precisely the closed, bounded intervals.)

If you want to find the absolute min and max, this is now a Lagrange multiplier problem, where you want to maximize $f$, subject to the constraint given by the ellipse equation. In general, these are easy to set up and harder to follow through on. There is only one constraint, so we can set $g(x,y)=3x^2+3y^2-2xy+4x+4y$, and then write down the system of equations:

$$\begin{cases} f_x(x,y)=\lambda g_x(x,y) \\ f_y(x,y)=\lambda g_y(x,y)\\ g(x,y)=4 \end{cases}$$

and hope for a tidy solution.

Answer 2

You say that the only critical point of $f(x,y)$ is $(0,0)$. Yes, it is, if we look at the domain $\mathbb R^2$.

But your question is about your $f$ on some one-dimensional subset of $\mathbb R^2$ and that subset happens to be an ellipse.

On that ellipse $x$ and $y$ are not at all independent but their relationship to each other is strictly determined by the equation defining your ellipse.

So, what happens here is that this function of yours will map your ellipse onto some segment $[a,b]$ in $\mathbb R$, so it will clearly have an absolute maximum and absolute minimum on that ellipse.

And they can be found.

I will not do that calculation here, this was written only to clarify what is not clear enough to you.

Of course that it would be much different if instead of $=$ we choose to take $\leq$ because then we would change very much the domain on which we observe our $f$, it would not be just an ellipse, but an ellipse + all of its interior.