Answer to Question #305438 in Math for Fresh

Question

Classify the critical points of f(x,y)= 4+x^3 + y^3 -3xy

Answer 1

$$f(x,y)= 4+x^3 + y^3 -3xy$$

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

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

Find the critical point(s)

$$f_x=0$$ $$f_y=0$$

Then

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

$$y=x^2$$

$$x^4-x=0$$

Critical points are $(0, 0), (1, 1)$

$$f_{xx}=6x$$

$$f_{xy}=f_{yx}=-3$$

$$f_{yy}=6y$$

$$D=\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=\begin{vmatrix} 6x & -3 \\ -3 & 6y \end{vmatrix}=36xy-9$$

Point $(0,0)$

$$D(0,0)=36(0)(0)-9=-9<0$$

Then $f(0,0)$ is not a local maximum or minimum. Point $(0,0)$ is a saddle point of $f.$

Point $(1,1)$

$$D(1,1)=36(1)(1)-9=27>0$$

$$f_{xx}(1,1)=6(1)=6>0$$

Then $f(1,1)=3$ is a local minimum.