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

The extreme f(x,y)=4x^2+1/4 y^3-2xy is Minimum(1/6 , 2/3),Saddle(0,0)

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extreme f(x,y)=4x^2+1/4 y^3-2xy

extreme

Minimum (\frac{1}{6}, \frac{2}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{6}, \frac{2}{3}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{3 y}{2}

D (x, y) = 12 y - 4

Check the sign of at each critical point

Check critical point

(\frac{1}{6}, \frac{2}{3}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{6}, \frac{2}{3}), Saddle (0, 0)

What is the extreme f(x,y)=4x^2+1/4 y^3-2xy ?
The extreme f(x,y)=4x^2+1/4 y^3-2xy is Minimum(1/6 , 2/3),Saddle(0,0)