Relative maximum and minimum values of the function $~f (x, y)~$

Question

Find the relative maximum / minimum values of the function $$f (x, y)= x^4 + y^4 -2x^2 +4xy -2y^2$$

I have calculated $$f_x=4x^3-4x+4y$$ $$f_{xx}=12x^2-4$$ $$f_y=4y^3+4x-4y$$ $$f_{yy}=12y^2-4$$ $$f_{xy}=4$$ for finding critical values $~f_x=0~$ and $~f_y=0~$ $$4x^3-4x+4y=0$$ $$\implies x^3-x+y=0$$ Similarly $$y^3+x-y=0$$

Now $~f_x=f_y ,~~~~y^3+x-y= x^3-x+y~~$ and got $~~(x+y)(x^2 +y^2-xy)=0~$

$x+y=0 ,~~~~ x=-y~$

Now check $$f_{xx}\cdot f_{yy} -(f_{xy})^2$$ $$= (12x^2-4)(12y^2-4)-4^2$$ $$=16[9x^2y^2-3x^2-3y^2]$$

Here I am stuck please help me

Answer 1

$$x^4+y^4-2x^2+4xy-2y^2=(x^2-2)^2+(y^2-2)^2+2(x+y)^2-8\geq-8.$$ The equality occurs for $x^2=y^2=2$ and $x+y=0,$ id est, for $(x,y)=(\sqrt2,-\sqrt2),$ for example,

which says that we got a minimal value.

The maximal value does not exist.

Try, $y=0$ and $x\rightarrow+\infty.$

Answer 2

Optimize $f (x, y)= x^4 + y^4 -2x^2 +4xy -2y^2$.

Once you find $x=-y$, you can plug it back to any of the equations: $f_x=0,f_y=0$, for example: $$f_x=4x^3-4x-4x=0 \Rightarrow x_1=0,x_{2,3}=\pm \sqrt2.$$ Now you can check the second order conditions: $$\Delta=f_{xx}f_{yy}-f_{xy}^2=(12x^2-4)(12y^2-4)-4^2;\\ (0,0): \Delta =0 \ \ \text{(inconclusive, but with extra checking it is found to be a saddle point)}\\ (\sqrt2,-\sqrt2): \Delta >0, f_{xx}=f_{yy}=20>0 \Rightarrow \text{min}\\ (-\sqrt2,\sqrt2): \Delta >0, f_{xx}=f_{yy}=20>0 \Rightarrow \text{min}$$ WA answer.