What is the extreme f(x,y)=3x^2y+y^3-6x^2-6y^2+4 ?

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

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extreme f(x,y)=3x^2y+y^3-6x^2-6y^2+4

extreme points

f (x, y) = 3 x^{2} y + y^{3} - 6 x^{2} - 6 y^{2} + 4

Minimum (0, 4), Maximum (0, 0), Saddle (2, 2), Saddle (- 2, 2)

Solution steps

Find the critical points:

(0, 4), (0, 0), (2, 2), (- 2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 12

D (x, y) = (6 y - 12)^{2} - 36 x^{2}

Check the sign of at each critical point

Check critical point

(0, 4) :

Minimum

Check critical point

(0, 0) :

Maximum

Check critical point

(2, 2) :

Saddle

Check critical point

(- 2, 2) :

Saddle

Minimum (0, 4), Maximum (0, 0), Saddle (2, 2), Saddle (- 2, 2)

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

f (x) = x^{2} y + y^{3} - 12 y

\frac{e ^{x}}{x + 1}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 2

f (x) = x (14 - 2 x) (15 - x)

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