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, y) = 2 x y^{4} - 5 x^{2} y^{3} - 3 x^{3} y

f (x) = x^{2} - 1, - 4 \leq x \leq 4

\frac{d}{d x} (6 x^{2} y - 2)

f (x) = x^{4} + 3 x^{2}

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