What is the extreme f(x,y)=x^2y+2xy^2-6xy-4 ?

The extreme f(x,y)=x^2y+2xy^2-6xy-4 is Minimum(2,1),Saddle(0,3)

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extreme f(x,y)=x^2y+2xy^2-6xy-4

extreme

Minimum (2, 1), Saddle (0, 3)

Solution steps

Find the critical points:

(2, 1), (0, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = - 4 x^{2} - 8 x y + 24 x - 16 y^{2} + 48 y - 36

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Check critical point

(0, 3) :

Saddle

Minimum (2, 1), Saddle (0, 3)

What is the extreme f(x,y)=x^2y+2xy^2-6xy-4 ?
The extreme f(x,y)=x^2y+2xy^2-6xy-4 is Minimum(2,1),Saddle(0,3)