What is the extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x ?

The extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x is Saddle(1,1),Maximum(-1,-1)

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extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x

extreme

Saddle (1, 1), Maximum (- 1, - 1)

Solution steps

Find the critical points:

(1, 1), (- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 48 x

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Saddle (1, 1), Maximum (- 1, - 1)

What is the extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x ?
The extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x is Saddle(1,1),Maximum(-1,-1)