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

The extreme f(x,y)=x^3+y^3-9xy is Saddle(0,0),Minimum(3,3)

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

extreme points

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

Saddle (0, 0), Minimum (3, 3)

Solution steps

Find the critical points:

(0, 0), (3, 3)

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

D (x, y) = 36 x y - 81

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(3, 3) :

Minimum

Saddle (0, 0), Minimum (3, 3)

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

x^{3} - 6 x^{2} + 5

f (x) = 3 - x, - 1 \leq x \leq 2

f (x) = x + 2 cos (x)

\frac{x ^{2}}{x + 1}

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