max and min of $f(x,y)=y^8-y^4x^6+x^4$

Question

The function is continuous in $\mathbb{R}^2$ and $f(x,y)=f(x,-y)=f(-x,y)=f(-x,-y)$. If I consider $f(x,0)=x^4$ for $x\rightarrow +\infty$ $f$ is not limited up so $\sup f(x,y)=+\infty$. But the origin is absolute min?

$f(x,x)=x^8-x^{10}+x^4\rightarrow -\infty$ if $x\rightarrow \infty$? so f is not limit down

Answer 1

$f'_x=-6x^5 y^4+4x^3;\;f'_y=8y^7-4y^3x^6$

They must be both zero

$ \left\{ \begin{array}{l} -2 x^3 (3 x^2 y^4-2)=0 \\ -4 y^3 \left(x^6-2 y^4\right)=0 \\ \end{array} \right. $

$(0,0)$ is a solution and then

$ \left\{ \begin{array}{l} 3 x^2 y^4-2=0 \\ x^6-2 y^4=0 \\ \end{array} \right. $

Solve for $y$ the first equation

$y^4=\dfrac{2}{3x^2}$

and plug in the second

$x^6-\dfrac{4}{3x^2}=0\to x^8=\dfrac{4}{3}\to x=\pm\sqrt[8]{\dfrac{4}{3}}$

and $y^4=\dfrac{2}{3\sqrt[4]\frac{4}{3}}=\dfrac{2}{3}\sqrt[4]{\dfrac{3}{4}}=\sqrt[4]{\dfrac{4}{27}}$

$y=\pm\sqrt[16]{\dfrac{4}{27}}$

The singular points are

$P_0(0,0);\;P_1\left(\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right);\;P_2\left(-\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right)$

$P_3\left(\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right);\;P_4\left(-\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right)$

To understand what is what we need second partial derivative test. Hessian matrix is:

$H(x,y)=\left( \begin{array}{ll} 12 x^2-30 x^4 y^4 & -24 x^5 y^3 \\ -24 x^5 y^3 & 56 y^6-12 x^6 y^2 \\ \end{array} \right)$

and its determinant

$D(x,y)=-24 x^2 y^2 \left(9 x^8 y^4+6 x^6+70 x^2 y^8-28 y^4\right)$

As the determinant contains all even powers of the variables and a coefficient negative, all $P_i,\;i=1,2,3,4$

Around zero $f(x,y)=x^4+O(x^5)+O(y^8)$ therefore $f$ is positive around the zero and the origin is a local minimum.

In the following graph we can see that $f(x,y)>0$ in a neighborhood of $(0,0)$

There is no global minimum as the function goes to $-\infty$ and no global maximum as the function goes to $+\infty$

Hope this helps

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Answer 2

$$f(0,y)=y^8\rightarrow+\infty$$ for $y\rightarrow\infty.$

In another hand, $$f(x,x)=-x^{10}+x^8+x^4\rightarrow-\infty$$ for $x\rightarrow\infty$,

which says that the maximum does not exist and the minimum does not exist.