Question

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

Answer 1

$$f'_x=3x^2-3y$$

$$f'_y=3y^2-3x$$

Find the critical point(s)

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

$$y=x^2$$ $$x^4-x=0$$

$$x(x-1)(x^2+x+1)=0$$ $$x=y^2$$

Critical point $(0, 0),$ critical point $(1, 1).$

Use Second Derivatives Test

$$f''_{xx}=6x$$

$$f''_{xy}=-3=f''_{yx}$$

$$f''_{yy}=6y$$

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

Critical point $(0, 0)$

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

$f(0,0)$ is not a local maximum or minimum.

Point $(0,0)$ is a saddle point of $f.$

Critical point $(1, 1)$

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

$f''_{xx}(1,1)=6(1)=6>0.$

$f(1,1)$ is a local minimum.