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4.4 More stationary points

Example.

Find the stationary points of

f⁢(x,y)=x⁢e-(x2+y2).f(x,y)=xe^{-(x^{2}+y^{2})}.

Solution. We have

∂⁡f∂⁡x=e-(x2+y2)-2⁢x2⁢e-(x2+y2)⁢and⁢∂⁡f∂⁡y=-2⁢x⁢y⁢e-(x2+y2).\frac{\partial f}{\partial x}=\,{e^{-(x^{2}+y^{2})}-2x^{2}e^{-(x^{2}+y^{2})}}% \;\;\mbox{and}\;\;\frac{\partial f}{\partial y}=\,{-2xye^{-(x^{2}+y^{2})}.}

Thus fx=fy=0f_{x}=f_{y}=0 if and only if 1-2⁢x2=-2⁢x⁢y=01-2x^{2}=-2xy=0. Therefore y= 0y=\,{0} and x=±12.x=\,{\pm\frac{1}{\sqrt{2}}.} Hence there are two stationary points: (±12,0)(\pm\frac{1}{\sqrt{2}},0) with corresponding values of f⁢(x,y)=±12⁢e-12.f(x,y)=\,{\pm\frac{1}{\sqrt{2}}e^{-\frac{1}{2}}.}