Answer to Question #217629 in Calculus for Njabsy

Question

1.consider the r^2-r function f defined by f(x,y)=x^2-6x+3y^2-y^3. (a) find all the critical points of f. (the function has two critical points). (b) use theorem 10.2.9 to determine the local extreme values and minimax values of f. also determine (by inspection) whether any of the local extrema are global extrema

Answer 1

(a)

$$f_x=2x-6$$

$$f_y=6y-3y^2$$

$$\begin{matrix} f_x=0\\ f_y=0 \end{matrix}$$

$$\begin{matrix} 2x-6=0\\ 6y-3y^2=0 \end{matrix}$$

$$\begin{matrix} x=3\\ y=0 \end{matrix}\ \ \ or\ \ \ \begin{matrix} x=3\\ y=2 \end{matrix}$$

Critical points: $(3, 0), (3, 2).$

(b)

$$f_{xx}=2>0$$

$$f_{xy}=f_{yx}=0$$

$$f_{yy}=6-6y$$

Point $(3, 0)$

$$D=\begin{vmatrix} 2 & 0 \\ 0 & 6-6(0) \end{vmatrix}=12>0$$

$D>0, f_{xx}>0.$ Point $(3,0)$ is a local minimum.

$$f(3, 0)=3^2-6(3)+3(0)^2-(0)^3=-9$$

Point $(3, 2)$

$$D=\begin{vmatrix} 2 & 0 \\ 0 & 6-6(2) \end{vmatrix}=-12<0$$

$D<0, f_{xx}>0.$ Point $(3,2)$ is a local maximum.

$$f(3, 2)=3^2-6(3)+3(2)^2-(2)^3=-5$$

$$f(0, 0)=0^2-6(0)+3(0)^2-(0)^3=0>-5$$

$$f(0, 10)=0^2-6(0)+3(10)^2-(10)^3=-700<-9$$

$$=-700<-9$$

Point $(3,0)$ is not a global extremum.

Point $(3,2)$ is not a global extremum.