What is the extreme f(x,y)=4x^3+y^3-12x-3y ?

The extreme f(x,y)=4x^3+y^3-12x-3y is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)

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extreme f(x,y)=4x^3+y^3-12x-3y

extreme points

f (x, y) = 4 x^{3} + y^{3} - 12 x - 3 y

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

Solution steps

Find the critical points:

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

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

D (x, y) = 144 x y

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

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

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

f (x) = 2 x^{2} - 4 x

f (x) = 12 x y - x^{3} - y^{3}

f (x, y) = 338 y^{2} + x^{2} - x^{2} y

f (x, y) = - x^{2} - y^{2} - \frac{1}{2} x

What is the extreme f(x,y)=4x^3+y^3-12x-3y ?
The extreme f(x,y)=4x^3+y^3-12x-3y is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)