What is the extreme f(x,y)=x^3-6xy+y^3+4 ?

The extreme f(x,y)=x^3-6xy+y^3+4 is Saddle(0,0),Minimum(2,2)

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extreme f(x,y)=x^3-6xy+y^3+4

extreme points

f (x, y) = x^{3} - 6 x y + y^{3} + 4

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)

- x^{3} + 6 x^{2} + x + 1

\frac{x ^{2} + 1}{( x - 1 ) ^{2}}

w (x, y) = x + x y

f (x) = 5 - x

f (x, y) = x^{2} + 3 y^{2} + 4 x - 9 y + 3

What is the extreme f(x,y)=x^3-6xy+y^3+4 ?
The extreme f(x,y)=x^3-6xy+y^3+4 is Saddle(0,0),Minimum(2,2)