Without Extreme Value Theorem, how do we find absolute extrema?

I have to find and classify the critical points of the following functions and then state which relative extrema are absolute extrema.

1. $$f(x,y) = x^3 - y^3 - 2xy + 6$$

2. $$f(x,y) = xy + 2x - \ln(x^2y), \ \text{in the first quadrant}$$

Does EVT apply? There are no bounds given.

I got the critical points from WA:

1. $(0,0), (-2/3,2/3)$ here

2. $(1/2,2)$ here

If EVT applies:

why?

Otherwise:

now what?

There are no bounds so how do I know if any local extrema is an absolute extrema?

I can use the second partial derivative test to classify them. Once I classify them, and I find that any of them are local extrema, how do I determine if any of those local extrema are absolute extrema?

Btw, for the second function, I don't think $x,y\ge0$ counts as satisfying assumptions of multivariable EVT.

What would it even mean to have $f(x,0)$ or $f(0,y)$ considering $\ln(x^2y)$ wouldn't be defined for $x=0$ or $y=0$?

Actually, would I have to show instead that the local extrema are not absolute extrema? Do I do that by finding another point that gives a lower or higher function value (depending on whether it's a local min or max)?

• In the first one you have to study the behavior at $\infty$ because this (loosely speaking) is the "boundary" of your domain, and so if there is an extremal point which is not in the interior, you will find it there. In the second you need to do the same, but this time restricting yourself to the first quadrant, and moreover studying the behavior on the Axes... – b00n heT May 14 '16 at 9:40
• @b00nheT Ah so for the second one the bounds are $x=0$ and $y=0$? then i just compare $f(1/2,2)$ with whatever I get from $f(0,y)$ and $f(x,0)$? So how do we know that there is an abs max and abs min? EVT applies or doesn't apply? For the first one, what? $f$ approaches $\infty$ or $-\infty$ if $x$ xor $y$ approaches $\infty$...right? – BCLC May 14 '16 at 12:51

A quick way to check is using graph. For the 1st problem, the graph looks like below. So both critical points are not extrema.

For the 2nd problem, the graph looks like below. So very likely, it is a minimum point. This is one quick way to make decision in daily work but by no means a mathematical proof. To prove it, you need calculate it Hessian matrix.

• Hessian matrices tell you whether or not a LOCAL extrema is an ABSOLUTE extrema? Are you sure about that? Please give me a reference. I can't find on Wikipedia – BCLC May 14 '16 at 16:24
• For unconstrained problem, Hessian matrix, like second order derivative in 1D, will tell you whether it is locally concave or convex, which leads to make your conclusion. I don't have website reference by hand, unfortunately. – user115350 May 14 '16 at 20:35
• how does it lead to the conclusion? According to stewart book I think rules in 1D second order don't apply to 2D order. Something like one critical point being loc min implies abs min in 1D but it doesn't hold true in 2D – BCLC May 14 '16 at 20:41
• where specifically is that? – BCLC May 15 '16 at 7:31

Here is another try by using the second function. $$f(x,y)=xy+2x-\ln (x^2y)$$

$$f_x=y+2-\frac2x$$ $$f_y=x-\frac1y$$ Thus, the critical point can be found by letting $$f_x=0$$ $$f_y=0$$ This gives $$x=0.5$$ $$y=2$$

Second derivative $$f_{xx}=\frac2{x^2}$$ $$f_{xy}=1$$ $$f_{yx}=1$$ $$f_{yy}=\frac1{y^2}$$

At critical point $(0.5, 2)$, Hessian matrix becomes, $$\left( \begin{matrix} 8 & 1 \\ 1 & 0.25 \\ \end{matrix} \right )$$

The eigenvalues are 8.12 and 0.12. Both positive. So this is local minimum.

If you don't want linear algebra. There is another way to look at this. You can draw an arbitrary line through critical point $(0.5, 2)$. $$y-2=\alpha (x-0.5)$$

Along the line, you can simplify the function to a single variable function.

$$f(x)=xy(x)+2x-\ln (x^2y(x))$$

The critical point is when $$\frac {df}{dx}=0$$

Derivative is $$\frac {df}{dx}=y+\alpha x+ 2 -\frac{y+\alpha x}{xy}$$

At $(0.5, 2)$, with any $\alpha$ $$\frac {df}{dx}=y+\alpha x+ 2 -\frac{y+\alpha x}{xy}=0$$

This verify the critical point.

Second derivative is $$\frac {d^2f}{dx^2}=2\alpha -\frac{3\alpha (xy)- (2y+\alpha x)(y+\alpha x)}{(xy)^2}$$

At $(0.5, 2)$, for any $\alpha$ $$\frac {d^2f}{dx^2}=4+(0.5 \alpha +2)^2 \gt 0$$

This means, the curve is convex for any sections. There, it is a minimum. You don't want to see the curve is convex for one section and concave for another section like saddle point.