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

The extreme f(x,y)=-3x^2+6x+12xy-3y^2+2y^3+6 is Saddle(-1,-1),Maximum(-3,-2)

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

extreme points

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

Saddle (- 1, - 1), Maximum (- 3, - 2)

Solution steps

Find the critical points:

(- 1, - 1), (- 3, - 2)

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

D (x, y) = - 72 y - 108

Check the sign of at each critical point

Check critical point

(- 1, - 1) :

Saddle

Check critical point

(- 3, - 2) :

Maximum

Saddle (- 1, - 1), Maximum (- 3, - 2)

f (x) = Y (x) = x^{3} - 3 x^{2} - 9 x + 20

f (x) = - 3 x^{\frac{2}{3}}, - 1 \leq x \leq 1

f (x) = x^{3} + 3 x^{2} - 24 x, - 1 \leq x \leq 3

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

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

What is the extreme f(x,y)=-3x^2+6x+12xy-3y^2+2y^3+6 ?
The extreme f(x,y)=-3x^2+6x+12xy-3y^2+2y^3+6 is Saddle(-1,-1),Maximum(-3,-2)