Answer to Question #223439 in Calculus for FRANCIS

Question

Find all the fixed points and classify each of them: f(x, y) = xye−x
2−y
2
.

Answer 1

$$f(x, y)=xye^{-x^2-y^2}$$

$$f_x=ye^{-x^2-y^2}-2x^2ye^{-x^2-y^2}$$

$$f_y=xe^{-x^2-y^2}-2xy^2e^{-x^2-y^2}$$

$$\begin{matrix} f_x=0 \\ f_y=0 \end{matrix}=>\begin{matrix} ye^{-x^2-y^2}-2x^2ye^{-x^2-y^2}=0 \\ xe^{-x^2-y^2}-2xy^2e^{-x^2-y^2}=0 \end{matrix}$$

$$\begin{matrix} y(1-2x^2)=0 \\ x(1-2y^2)=0 \end{matrix}$$

Critical points:

$$(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), (-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}), (0, 0),$$

$$(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), (\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$$

$$f_{xx}=-6xye^{-x^2-y^2}+4x^3ye^{-x^2-y^2}$$

$$f_{xy}=e^{-x^2-y^2}-2y^2e^{-x^2-y^2}-2x^2e^{-x^2-y^2}+4x^2y^2e^{-x^2-y^2}$$

$$f_{yy}=-6xye^{-x^2-y^2}+4xy^3e^{-x^2-y^2}$$

$(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})$

$$f_{xx}=-2e^{-1}<0$$

$$f_{xy}=0$$

$$f_{yy}=-2e^{-1}$$

$$\begin{vmatrix} -2e^{-1} & 0 \\ 0 & -2e^{-1} \end{vmatrix}=4e^{-2}>0$$

Point $(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})$ is a local maximum.

$(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$

$$f_{xx}=2e^{-1}>0$$

$$f_{xy}=0$$

$$f_{yy}=2e^{-1}$$

$$\begin{vmatrix} 2e^{-1} & 0 \\ 0 & 2e^{-1} \end{vmatrix}=4e^{-2}>0$$

Point $(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$ is a local minimum.

$(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})$

$$f_{xx}=2e^{-1}>0$$

$$f_{xy}=0$$

$$f_{yy}=2e^{-1}$$

$$\begin{vmatrix} 2e^{-1} & 0 \\ 0 & 2e^{-1} \end{vmatrix}=4e^{-2}>0$$

Point $(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})$ is a local minimum.

$(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$

$$f_{xx}=-2e^{-1}<0$$

$$f_{xy}=0$$

$$f_{yy}=-2e^{-1}$$

$$\begin{vmatrix} -2e^{-1} & 0 \\ 0 & -2e^{-1} \end{vmatrix}=4e^{-2}>0$$

Point $(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$ is a local maximum.

$(0,0)$

$$f_{xx}=0$$

$$f_{xy}=1$$

$$f_{yy}=0$$

$$\begin{vmatrix} 0 & 1\\ 1 & 0 \end{vmatrix}=-1<0$$

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