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

The extreme f(x,y)=2x^3+6xy+3y^2 is Minimum(1,-1),Saddle(0,0)

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

extreme points

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

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

Solution steps

Find the critical points:

(1, - 1), (0, 0)

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

D (x, y) = 72 x - 36

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Check critical point

(0, 0) :

Saddle

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

g (x, y) = x^{2} y^{3} + x^{4} y + 3 y^{2}

f (x, y) = - 2 x^{2} - y^{2} + 8 x + 10 y - 5 x y

f (x, y) = 4 x y^{2} - 2 x^{2} y - x

\frac{6 - 2 a}{30}

x^{2} y - 15 x y^{2} + 12 y

What is the extreme f(x,y)=2x^3+6xy+3y^2 ?
The extreme f(x,y)=2x^3+6xy+3y^2 is Minimum(1,-1),Saddle(0,0)