Question

Consider the R²-R function f defined by f(x,y)=x²+2y²-x²y.

Show that f has two saddle points

Answer 1

$$f_x=2x-2xy$$

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

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

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

If $x=0,$ then $y=0.$

If $y=1,$ then $x=-2$ or $x=2.$

Point (0,0), Point(-2, 1), Point(2, 1).

$$f_{xx}=2-2y, f_{yy}=4, f_{xy}=f_{yx}=-2x$$

$$D=\begin{vmatrix} 2-2y & -2x \\ -2x & 4 \end{vmatrix}=8-8y-4x^2$$

Point (0,0)

$$D=8-0-0=8>0$$

$$f_{xx}(0,0)=2>0$$

Then $f(0,0)$ is a local minimum.

Point (-2,1)

$$D=8-8-16=-16<0$$

Then $f(-2,1)$ is a saddle point.

Point (2,1)

$$D=8-8-16=-16<0$$

Then $f(2,1)$ is a saddle point.