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

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

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

extreme points

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

Saddle (0, 0), Minimum (\frac{1}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{3}, \frac{1}{3})

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

D (x, y) = 36 x y - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{3}, \frac{1}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{1}{3}, \frac{1}{3})

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

f (x) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 1

f (x) = x^{\frac{4}{3}} - x^{\frac{2}{3}}

5 x^{3} - 5 x^{2} - 5 x + 9

\frac{x ^{3}}{x ^{2} - 49}

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