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

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

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

extreme points

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

Maximum (0, 0), Saddle (2, - 2), Saddle (- 4, - 8)

Solution steps

Find the critical points:

(0, 0), (2, - 2), (- 4, - 8)

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

D (x, y) = - 6 (12 x - 6 y - 24) - 36 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Check critical point

(2, - 2) :

Saddle

Check critical point

(- 4, - 8) :

Saddle

Maximum (0, 0), Saddle (2, - 2), Saddle (- 4, - 8)

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