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

The extreme f(x,y)=4+x^3+y^3-3xy is Saddle(0,0),Minimum(1,1)

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

extreme points

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

Saddle (0, 0), Minimum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (1, 1)

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

\frac{x}{x ^{2} - 1}

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

f (x) = x \cdot ln (x)

f (x, y) = 2 x^{2} + 3 x y + y^{2} + 0 x + 5

What is the extreme f(x,y)=4+x^3+y^3-3xy ?
The extreme f(x,y)=4+x^3+y^3-3xy is Saddle(0,0),Minimum(1,1)