What is the extreme f(x,y)=2x^3-9x^2+12x+2y^3+3y^2 ?

The extreme f(x,y)=2x^3-9x^2+12x+2y^3+3y^2 is Saddle(1,0),Maximum(1,-1),Minimum(2,0),Saddle(2,-1)

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extreme f(x,y)=2x^3-9x^2+12x+2y^3+3y^2

extreme points

f (x, y) = 2 x^{3} - 9 x^{2} + 12 x + 2 y^{3} + 3 y^{2}

Saddle (1, 0), Maximum (1, - 1), Minimum (2, 0), Saddle (2, - 1)

Solution steps

Find the critical points:

(1, 0), (1, - 1), (2, 0), (2, - 1)

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

D (x, y) = (12 y + 6) (12 x - 18)

Check the sign of at each critical point

Check critical point

(1, 0) :

Saddle

Check critical point

(1, - 1) :

Maximum

Check critical point

(2, 0) :

Minimum

Check critical point

(2, - 1) :

Saddle

Saddle (1, 0), Maximum (1, - 1), Minimum (2, 0), Saddle (2, - 1)

f (x) = x 6 - x^{2}

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

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

f (x, y) = 5 x^{2} - 3 y^{2} + 4 x y - 32 x + 10 y

f (x, y) = - 3 x^{2} - 2 y^{2} + 3 x - 4 y + 5

What is the extreme f(x,y)=2x^3-9x^2+12x+2y^3+3y^2 ?
The extreme f(x,y)=2x^3-9x^2+12x+2y^3+3y^2 is Saddle(1,0),Maximum(1,-1),Minimum(2,0),Saddle(2,-1)