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

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

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

extreme points

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

Saddle (0, 0), Maximum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

D (x, y) = 72 y - 72

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Maximum

Saddle (0, 0), Maximum (2, 2)

f (x, y) = \frac{3}{2} x^{2} + x^{3} + 3 y^{2}

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

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

f (x) = x^{3} + 2 x^{2} - 15 x - 20

f (x, y) = x^{3} + 3 x y^{2} - 15 x - 12 y

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