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

The extreme f(x,y)=x^3-3xy+3y^2 is Minimum(1/2 , 1/4),Saddle(0,0)

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

extreme points

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

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, \frac{1}{4}), (0, 0)

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

D (x, y) = 36 x - 9

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, \frac{1}{4}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

f (x, y) = 7 e^{y} - 7 y e^{x}

f (x) = x^{4} - 50 x^{2} + 3

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

f (x) = 4 x e^{3 x}

f (x) = 6 x^{\frac{1}{6}} y^{\frac{1}{6}} - x - y

What is the extreme f(x,y)=x^3-3xy+3y^2 ?
The extreme f(x,y)=x^3-3xy+3y^2 is Minimum(1/2 , 1/4),Saddle(0,0)