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

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

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

extreme points

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

Saddle (0, 0), Maximum (- 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) :

Maximum

Saddle (0, 0), Maximum (- 1, 1)

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

f (x) = (x - 3) (x - 7)^{3} + 12, (1, 4)

f (x, y) = 2 x^{3} + y^{2} - 6 x y - 23

f (x) = \frac{3 x ^{2}}{x ^{2} - 9}

x^{2} + y^{2} - 8 y + 16

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