What is the extreme f(x,y)=9x^3+(y^3)/3-4xy ?

The extreme f(x,y)=9x^3+(y^3)/3-4xy is Saddle(0,0),Minimum(4/9 , 4/3)

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extreme f(x,y)=9x^3+(y^3)/3-4xy

extreme points

f (x, y) = 9 x^{3} + \frac{y ^{3}}{3} - 4 x y

Saddle (0, 0), Minimum (\frac{4}{9}, \frac{4}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{4}{9}, \frac{4}{3})

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

D (x, y) = 108 x y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{4}{9}, \frac{4}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{4}{9}, \frac{4}{3})

f (x) = x + y

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

f (x) = x^{4} - 4 x^{3} - 8 x^{2}

f (x, y) = x^{3} ln (y^{2} + 1) - 2 y

f (x) = (\frac{x}{5})^{1} \cdot (\frac{y}{5})^{x - 1}

What is the extreme f(x,y)=9x^3+(y^3)/3-4xy ?
The extreme f(x,y)=9x^3+(y^3)/3-4xy is Saddle(0,0),Minimum(4/9 , 4/3)