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Critical point classification III

mark

Find and classify the critical points of

f(x,y) = x^3 - 3xy - y^2 - y.

mearing

Partial derivatives:

f_{x}=3x^2-3y

f_{y}=-2y-3x-1

Solve:

3x^2-3y=0

and

-2y-3x-1=0

Solutions:

(-1,1) , (-\frac{1}{2},\frac{1}{4})

Second derivatives

f_{xx}=6x

f_{xy}/f_{yx}=-3

f_{yy}=-2

Check D(x_{0},y_{0}):

D(-1,1)=6(-1)*(-2)-(-3)^2

D(-1,1)=3

D(-\frac{1}{2},\frac{1}{4})=6(-\frac{1}{2})*(-2)-(-3)^2

D(-\frac{1}{2},\frac{1}{4})=-3

Saddle point at \cancel{ (-1,1)} and maximum at \cancel{(-\frac{1}{2},\frac{1}{4})}
Saddle point at (-\frac{1}{2},\frac{1}{4}) and maximum at (-1,1)

mark

@mearing This looks great! But something still seems a bit off. I think you’ve got the right critical points but my graphical experiment below indicate that your classification might not be correct.