How to prove that a function $f(x,y)$ has a maximum or minimum value?

Question

I am given a function $f(x,y)=\frac{1-2xy}{x^2+y^2}$ and I am supposed to prove whether or not the function has a maximum or minimum value for $(x,y)\neq (0,0)$.

I've done lots of exercises regarding critical points, but I can't seem to get the "trick" for this one.

I have tried the normal approach of taking $\nabla f(x,y) = 0$ to find values for $x$ and $y$. However the calculation quickly develops into substituting a 2nd degree equation of either $x$ or $y$ into the other equation derived from $\nabla f(x,y) = 0$. Because of this I get a sense there is a "trick" to solving this, but I can't seem to find it.

This is how my $f_x$ and $f_y$ looks like, and the equations I end up solving using the 2nd degree equation solving formula:

$$f_x = \frac{2(x^2y - x - y^3)}{(x^2 + y^2)^2} = 0$$ $$f_y = \frac{2(y^2x - y - x^3)}{(x^2 + y^2)^2} = 0$$

$$x^2y - x - y^3 = 0$$ $$y^2x - y - x^3 = 0$$

So my question is really, whats the trick? or is this actually just a "brute force" solving exercise?

Answer 1

Multiply $x^2y - x - y^3 = 0$ by $x$ and multiply $y^2x - y - x^3 = 0$ by $y$ and add the resulting equations. Then you get $x^2+y^2=0$.

Conclusion ?

Answer 2

proving that $$\frac{1-2xy}{x^2+y^2}\geq -1$$ this is equivalent to $$-1\le (x-y)^2$$ there isn't a maximum multiplying by $$x^2+y^2\ne 0$$ we get $$1-2xy\geq -x^2-y^2$$ and this is $$1\geq -x^2-y^2+2xy$$ and this is $$1\geq -(x-y)^2$$

Answer 3

If you rewrite it as polar coordinate, we have

$$\frac{1}{r^2}-\sin(2\theta)$$

Hence fixing $\theta$ and $r$ reduces, the function become arbitrarily large.

Also from the polar coordimate, we know that the function is at least $\frac1{r^2}-1$ but we can't attain $-1$ though we get arbitrarily close to it.

Answer 4

For $x=0$ and $y\rightarrow0$ we see that $f\rightarrow+\infty$, which says that the maximum does not exist.

In another hand, $f(x,y)>-\frac{-2xy}{x^2+y^2}\geq-1$ because the last inequality it's $(x-y)^2\geq0.$

Also, we see that for $x=y\rightarrow+\infty$ we have $f\rightarrow-1,$

which gives infimum $-1$ and the minimum does not exist.