What is the extreme f(x,y)=6x^2y+2y^3-12x^2-12y^2+8 ?

The extreme f(x,y)=6x^2y+2y^3-12x^2-12y^2+8 is Minimum(0,4),Maximum(0,0),Saddle(2,2),Saddle(-2,2)

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extreme f(x,y)=6x^2y+2y^3-12x^2-12y^2+8

extreme points

f (x, y) = 6 x^{2} y + 2 y^{3} - 12 x^{2} - 12 y^{2} + 8

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}} = 12 y - 24

D (x, y) = (12 y - 24)^{2} - 144 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) = (\frac{x ^{3} + x ^{2} - 12 x}{3})

f (x) = - sin (x)

f (x) = \frac{x}{ln ( x )} a t [2.5]

f (x) = (x + 5)^{3}

f (x) = x^{6} (1 - x)^{3}

What is the extreme f(x,y)=6x^2y+2y^3-12x^2-12y^2+8 ?
The extreme f(x,y)=6x^2y+2y^3-12x^2-12y^2+8 is Minimum(0,4),Maximum(0,0),Saddle(2,2),Saddle(-2,2)