What is the extreme f(x,y)= 1/3 y^3+x^2-2xy-15y+12 ?

The extreme f(x,y)= 1/3 y^3+x^2-2xy-15y+12 is Minimum(5,5),Saddle(-3,-3)

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extreme f(x,y)= 1/3 y^3+x^2-2xy-15y+12

extreme points

f (x, y) = \frac{1}{3} y^{3} + x^{2} - 2 x y - 15 y + 12

Minimum (5, 5), Saddle (- 3, - 3)

Solution steps

Find the critical points:

(5, 5), (- 3, - 3)

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

D (x, y) = 4 y - 4

Check the sign of at each critical point

Check critical point

(5, 5) :

Minimum

Check critical point

(- 3, - 3) :

Saddle

Minimum (5, 5), Saddle (- 3, - 3)

F (x, y) = y^{3} + 3 x^{2} y - 6 x^{2} - 6 y^{2} + 2

f (x, y) = - x^{2} - 4 y^{2} + 6 x + 32 y - 76

f (x, y) = 3 x y^{2} - x^{3} - 15 x + 36 y + 9

f (x, y) = \frac{ln ( x )}{y - ∣ x ∣}

What is the extreme f(x,y)= 1/3 y^3+x^2-2xy-15y+12 ?
The extreme f(x,y)= 1/3 y^3+x^2-2xy-15y+12 is Minimum(5,5),Saddle(-3,-3)