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

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

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

extreme points

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

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)

y = f (x) = - x (x - 1)^{5}

f (x) = \frac{12}{x} + 3 π x^{2}

f (x) = x^{\frac{4}{7}} (\frac{11}{4} + x)

0.5 x^{2} - 2 x - 7

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

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