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)

x^{4} + y^{4} - 4 x y + 1

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

f (x) = x^{2} - y^{2}

f (x) = x - 2 + 4 - x

f (x) = x + cos (x)

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)