What is the extreme f(x,y)=(x^3)/3-(y^3)/3+2xy ?

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

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extreme f(x,y)=(x^3)/3-(y^3)/3+2xy

extreme points

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

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

Solution steps

Find the critical points:

(0, 0), (2, - 2)

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

D (x, y) = - 4 x y - 4

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) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x

14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

f (x) = - x^{3} + 12 x - 8

f (x) = x^{2} - 11, - 2 \leq x \leq 3

f (x y) = x^{3} + y^{2} - 6 x y + 9 x + 5 y + 2

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