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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 36 y - 9

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(0, 0) :

Saddle

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

f (x) = a^{3} y = x^{3} (4 a - 3 x)

f (x) = \frac{10}{3} x^{3} - 11 x^{2} + 12 x - 24

y = x + sin (x)

x^{3} - 3 x^{2} - 9 x + 5

\frac{x}{x ^{2} + 1}

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