What is the extreme f(x,y)=x^3+y^3-3xy+1 ?

The extreme f(x,y)=x^3+y^3-3xy+1 is Saddle(0,0),Minimum(1,1)

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extreme f(x,y)=x^3+y^3-3xy+1

extreme points

f (x, y) = x^{3} + y^{3} - 3 x y + 1

Saddle (0, 0), Minimum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (1, 1)

f (x) = x^{2} + 36

y = 3 x^{2} + 12 x

f (x) = - 2 x^{3} + 12 x^{2} + 18

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

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

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