What is the extreme f(x,y)=x^3+y^3-6xy+27 ?

The extreme f(x,y)=x^3+y^3-6xy+27 is Saddle(0,0),Minimum(2,2)

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extreme f(x,y)=x^3+y^3-6xy+27

extreme points

f (x, y) = x^{3} + y^{3} - 6 x y + 27

Saddle (0, 0), Minimum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Minimum

Saddle (0, 0), Minimum (2, 2)

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

f (x) = x^{4} - 2 x^{3} + 1

\frac{2 x}{x ^{2} - 9}

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

f (x, y) = x^{2} - 2 x + y^{2} + 2 y + 1

What is the extreme f(x,y)=x^3+y^3-6xy+27 ?
The extreme f(x,y)=x^3+y^3-6xy+27 is Saddle(0,0),Minimum(2,2)