What is the extreme f(x,y)=2x^3-3x^2+2y^3-6y^2 ?

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

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extreme f(x,y)=2x^3-3x^2+2y^3-6y^2

extreme points

f (x, y) = 2 x^{3} - 3 x^{2} + 2 y^{3} - 6 y^{2}

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

Solution steps

Find the critical points:

(0, 2), (0, 0), (1, 2), (1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y - 12

D (x, y) = (12 x - 6) (12 y - 12)

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(0, 0) :

Maximum

Check critical point

(1, 2) :

Minimum

Check critical point

(1, 0) :

Saddle

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

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