What is the extreme f(x,y)=x^2y+y^3-12y ?

The extreme f(x,y)=x^2y+y^3-12y is Minimum(0,2),Maximum(0,-2),Saddle(2sqrt(3),0),Saddle(-2sqrt(3),0)

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extreme f(x,y)=x^2y+y^3-12y

extreme points

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

Minimum (0, 2), Maximum (0, - 2), Saddle (23, 0), Saddle (- 23, 0)

Solution steps

Find the critical points:

(0, 2), (0, - 2), (23, 0), (- 23, 0)

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

D (x, y) = 12 y^{2} - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, - 2) :

Maximum

Check critical point

(23, 0) :

Saddle

Check critical point

(- 23, 0) :

Saddle

Minimum (0, 2), Maximum (0, - 2), Saddle (23, 0), Saddle (- 23, 0)

f (x, y) = \frac{x y}{1 + e ^{x}}

f (x, y) = 5 + 9 - x^{2} - y^{2}

f (x, y) = 2 x + y + \frac{2}{x} + \frac{4}{y}

f (x, y) = - x - y + 2

f (x, y) = 18 x y - 3 x^{2} y - 2 x y^{2}

What is the extreme f(x,y)=x^2y+y^3-12y ?
The extreme f(x,y)=x^2y+y^3-12y is Minimum(0,2),Maximum(0,-2),Saddle(2sqrt(3),0),Saddle(-2sqrt(3),0)