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

The extreme f(x,y)=3x^2-xy+2y^3 is Minimum(1/216 , 1/36),Saddle(0,0)

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

extreme points

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

Minimum (\frac{1}{216}, \frac{1}{36}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{216}, \frac{1}{36}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y

D (x, y) = 72 y - 1

Check the sign of at each critical point

Check critical point

(\frac{1}{216}, \frac{1}{36}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{216}, \frac{1}{36}), Saddle (0, 0)

What is the extreme f(x,y)=3x^2-xy+2y^3 ?
The extreme f(x,y)=3x^2-xy+2y^3 is Minimum(1/216 , 1/36),Saddle(0,0)