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

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

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extreme f(xy)=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) = x^{2} ln (x), \frac{1}{3} \leq x \leq \frac{5}{6}

f (x) = - x^{4} + 6 x^{2}

f (x) = \frac{2}{x} + x^{2}

f (x) = t 9 - t^{2}

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

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