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

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

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

extreme

Minimum (0, 0), Saddle (0, 2), Saddle (- \frac{1}{3}, 0), Maximum (- \frac{1}{3}, 2)

Solution steps

Find the critical points:

(0, 0), (0, 2), (- \frac{1}{3}, 0), (- \frac{1}{3}, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 y + 6

D (x, y) = (24 x + 4) (- 6 y + 6)

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, 2) :

Saddle

Check critical point

(- \frac{1}{3}, 0) :

Saddle

Check critical point

(- \frac{1}{3}, 2) :

Maximum

Minimum (0, 0), Saddle (0, 2), Saddle (- \frac{1}{3}, 0), Maximum (- \frac{1}{3}, 2)

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