What is the extreme x^3+y^3-2xy ?

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

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extreme x^3+y^3-2xy

extreme points

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

Saddle (0, 0), Minimum (\frac{2}{3}, \frac{2}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{2}{3}, \frac{2}{3})

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

D (x, y) = 36 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{2}{3}, \frac{2}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{2}{3}, \frac{2}{3})

f (x, y) = x^{3} + y^{3} + 6 x^{2} - 9 y^{2} - 4

f (x) = \frac{1}{9} (x^{4} - 4 x^{3})

f (x) = x^{3} + 2 x^{2} - x - 10

f (x) = \frac{x ^{3} + 1}{x}

f (x) = - 2 x^{2} + 12 x

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