What is the extreme f(x,y)= 1/3 x^3+1/3 y^3-xy+4 ?

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

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extreme f(x,y)= 1/3 x^3+1/3 y^3-xy+4

extreme points

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x y + 4

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

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 4 x y - 1

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^{2} - x^{2}

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

T (x, y) = (x + 3 y, - x + 5 y)

f (x) = x (12 - 2 x)^{2}

f (x, y) = 8 x^{3} + y^{3} + 6 x y

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