Answer to Question #223439 in Calculus for FRANCIS

Question #223439
Find all the fixed points and classify each of them: f(x, y) = xye−x
2−y
2
.
1
Expert's answer
2021-08-05T13:58:58-0400
f(x,y)=xyex2y2f(x, y)=xye^{-x^2-y^2}

fx=yex2y22x2yex2y2f_x=ye^{-x^2-y^2}-2x^2ye^{-x^2-y^2}

fy=xex2y22xy2ex2y2f_y=xe^{-x^2-y^2}-2xy^2e^{-x^2-y^2}

fx=0fy=0=>yex2y22x2yex2y2=0xex2y22xy2ex2y2=0\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}

y(12x2)=0x(12y2)=0\begin{matrix} y(1-2x^2)=0 \\ x(1-2y^2)=0 \end{matrix}

Critical points:

(22,22),(22,22),(0,0),(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), (-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}), (0, 0),

(22,22),(22,22)(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), (\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})

fxx=6xyex2y2+4x3yex2y2f_{xx}=-6xye^{-x^2-y^2}+4x^3ye^{-x^2-y^2}

fxy=ex2y22y2ex2y22x2ex2y2+4x2y2ex2y2f_{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}

fyy=6xyex2y2+4xy3ex2y2f_{yy}=-6xye^{-x^2-y^2}+4xy^3e^{-x^2-y^2}

(22,22)(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})


fxx=2e1<0f_{xx}=-2e^{-1}<0

fxy=0f_{xy}=0

fyy=2e1f_{yy}=-2e^{-1}

2e1002e1=4e2>0\begin{vmatrix} -2e^{-1} & 0 \\ 0 & -2e^{-1} \end{vmatrix}=4e^{-2}>0

Point (22,22)(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}) is a local maximum.



(22,22)(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})


fxx=2e1>0f_{xx}=2e^{-1}>0

fxy=0f_{xy}=0

fyy=2e1f_{yy}=2e^{-1}

2e1002e1=4e2>0\begin{vmatrix} 2e^{-1} & 0 \\ 0 & 2e^{-1} \end{vmatrix}=4e^{-2}>0

Point (22,22)(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}) is a local minimum.



(22,22)(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})

fxx=2e1>0f_{xx}=2e^{-1}>0

fxy=0f_{xy}=0

fyy=2e1f_{yy}=2e^{-1}

2e1002e1=4e2>0\begin{vmatrix} 2e^{-1} & 0 \\ 0 & 2e^{-1} \end{vmatrix}=4e^{-2}>0

Point (22,22)(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}) is a local minimum.


(22,22)(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})


fxx=2e1<0f_{xx}=-2e^{-1}<0

fxy=0f_{xy}=0

fyy=2e1f_{yy}=-2e^{-1}

2e1002e1=4e2>0\begin{vmatrix} -2e^{-1} & 0 \\ 0 & -2e^{-1} \end{vmatrix}=4e^{-2}>0

Point (22,22)(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}) is a local maximum.



(0,0)(0,0)


fxx=0f_{xx}=0

fxy=1f_{xy}=1

fyy=0f_{yy}=0

0110=1<0\begin{vmatrix} 0 & 1\\ 1 & 0 \end{vmatrix}=-1<0

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




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