What is the extreme f(x,y)=x^3+y^3-6x^2-3/2 y^2+9x-1 ?

The extreme f(x,y)=x^3+y^3-6x^2-3/2 y^2+9x-1 is Saddle(1,1),Maximum(1,0),Minimum(3,1),Saddle(3,0)

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extreme f(x,y)=x^3+y^3-6x^2-3/2 y^2+9x-1

extreme points

f (x, y) = x^{3} + y^{3} - 6 x^{2} - \frac{3}{2} y^{2} + 9 x - 1

Saddle (1, 1), Maximum (1, 0), Minimum (3, 1), Saddle (3, 0)

Solution steps

Find the critical points:

(1, 1), (1, 0), (3, 1), (3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 3

D (x, y) = (6 y - 3) (6 x - 12)

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(1, 0) :

Maximum

Check critical point

(3, 1) :

Minimum

Check critical point

(3, 0) :

Saddle

Saddle (1, 1), Maximum (1, 0), Minimum (3, 1), Saddle (3, 0)

f (x) = \frac{x ^{3}}{3} - 5 x^{2} + 16 x - 100

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

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

f (x, y) = 3 x + 5 y

f (x) = x y (1 - x - y)

What is the extreme f(x,y)=x^3+y^3-6x^2-3/2 y^2+9x-1 ?
The extreme f(x,y)=x^3+y^3-6x^2-3/2 y^2+9x-1 is Saddle(1,1),Maximum(1,0),Minimum(3,1),Saddle(3,0)