What is the extreme f(x,y)=x^3+6xy+3y^2 ?

The extreme f(x,y)=x^3+6xy+3y^2 is Minimum(2,-2),Saddle(0,0)

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extreme f(x,y)=x^3+6xy+3y^2

extreme points

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

Minimum (2, - 2), Saddle (0, 0)

Solution steps

Find the critical points:

(2, - 2), (0, 0)

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

D (x, y) = 36 x - 36

Check the sign of at each critical point

Check critical point

(2, - 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (2, - 2), Saddle (0, 0)

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

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

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 2

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

f (x, y) = e^{x} - x e^{y}

What is the extreme f(x,y)=x^3+6xy+3y^2 ?
The extreme f(x,y)=x^3+6xy+3y^2 is Minimum(2,-2),Saddle(0,0)