What are the extrema and saddle points of #f(x,y) = e^y(y^2-x^2)#?

1 Answer
Jul 2, 2016

Answer:

#{0,0}# saddle point
#{0,-2}# local maximum

Explanation:

#f(x,y) = e^y(y^2-x^2)#

so the sationary points are determined by solving

#grad f(x,y) = vec 0#

or

#{ (-2 e^y x = 0), (2 e^y y + e^y (-x^2 + y^2) = 0) :}#

giving two solutions

#((x=0,y=0),(x=0,y=-2))#

Those points are qualified using

#H = grad(grad f(x,y))#

or

#H =((-2 e^y, -2 e^y x),(-2 e^y x, 2 e^y + 4 e^y y + e^y (-x^2 + y^2))) #

so

#H(0,0) = ((-2, 0),(0, 2))# has eigenvalues #{-2,2}#. This result qualifies point #{0,0}# as a saddle point.

#H(0,-2)=((-2/e^2, 0),(0, -2/e^2))# has eigenvalues #{-2/e^2, -2/e^2}#. This result qualifies point #{0,-2}# as a local maximum.

Attached the #f(x,y)# contour map near the points of interest

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