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

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Answer 1 · 2016-07-02T17:31:48

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

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

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

enter image source here

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