Maximum and minimum absolute of a function $(x,y)$

Question

I want to know the maximum and minimum absolutes values of this function:

$\ f(x,y)= 4x^2 + 9y^2 - x^2y^2 $

$\nabla f(x,y)=(8x-2xy^2,18y-2yx^2) $

I find these critical points:

$\ (0,0);(3,2);(-3,2);(3,-2);(-3;-2) $

In$\ (0,0)$ the Hessian critirium work,$\ (0,0)$ its a local minimum

But wait:

for $\ (3,2)$: $$ \begin{bmatrix} 0 & -24 \\ -24 & 0 \\ \end{bmatrix} $$

So in this case (and in the other points) the Hessian criterium doesn't work.

How I can demonstrate than $\ (0,0)$ es a absolute minimum of the function, and what about the other points? Are they saddle points? Max, Min?

Thanks very much, I'm sorry for my bad English.

Here in Argentina, the university is free, but English lessons not.

Answer 1

Hint: Test for Extrema

Let $f$ be a function of two variables that has continuous second partial derivatives on a rectangular region $Q$ and let:

$$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$$

for every $(x,y)$ in $Q$.

If $(a,b)$ is in $Q$ and $f_x(a,b) = 0, f_y(a,b) = 0$, then:

• (i) $f(a, b)$ is a local maximum of $f$ if $g(a,b) \gt 0$ and $f_{xx}(a, b) \lt 0$.
• (ii) $f(a, b)$ is a local minimum of $f$ if $g(a,b) \gt 0$ and $f_{xx}(a, b) \gt 0$.
• (iii) $f(a,b)$ is not an extremum of $f$ if $g(a,b) \lt 0$

We are given:

$$f(x,y)= 4x^2 + 9y^2 - x^2y^2$$

The critical points (you did great) are:

$$(0,0);(3,2);(-3,2);(3,-2);(-3,-2)$$

We have:

• $f_x = 8x - 2xy^2, f_{xx} = 8 - 2y^2$
• $f_y = 18y - 2x^2y, f_{yy} = 18 - 2x^2$
• $f_{xy} = -4xy$

Now evaluate each of the critical points using the above definition.

Also, it helps to sometimes plot the function and we have:

Examples:

• At $(x, y) = (0, 0)$, we have $g(0,0) \gt 0$ and $f_{xx} \gt 0 \rightarrow$ a local minimum
• At $(x, y) = (3, 2)$, we have $g(3, 2) \lt 0 \rightarrow$ not an extremum
• Repeat for the remaining critical points

Update:

The Hessian determinant is given by:

$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$

If you are using the Hessian, there are four conditions you need to test:

• (01) $f_{xx} \gt 0$ and $\det H \gt 0 \rightarrow$ local minimum
• (02) $f_{xx} \lt 0$ and $\det H \gt 0 \rightarrow$ local maximum
• (03) $\det(H) \lt 0 \rightarrow$ saddle point
• (04) $\det(H) = 0 \rightarrow$ no statement can be made using this approach

So, for each critical point that you found, can you now classify it using the criteria above and your approach?

Can you compare this to the approach above?

Can you look at other criteria to help distinguish the critical points as alluded to in the comments?

Answer 2

Checking up the analysis of Amzoti, I figered out something:

$\ (3,2);(-3,2);(3,-2);(-3;-2) $ Are not extreme points.

$\ (0,0)$ Is an extreme? Lets check:

if $\ (0,0) $ is minimum, then:

$\ f(x,y) > f(0,0) $

Then

$\ f(x,0)= 4x^2 (0,0) \ is \ minimum $

$\ f(0,7)= 9y^2 (0,0) \ is \ minimum $

$\ f(x,x)= 11x^2-x^4 = h(x)$

$\ h(x)= 11x^2-x^4$

$\ h'(x)= x(22-4x^2) \\ h'(0)=0 \\ h'(\sqrt{5.5})=0 \\ h'(-\sqrt{5.5})=0 $

$\ h''(x)= 22-12x^2 \\ h''(0)= 22 $
$\ h''(\sqrt{5.5})= -44 \\ h''(-\sqrt{5.5})= -44 $

$\ (0,0) $ is not Minimum neither maximum.

$\ f(10,10)<F(0,0) $