What is the extreme f(x)=x^3-8y^3+4xy-8 ?

The extreme f(x)=x^3-8y^3+4xy-8 is Saddle(0,0),Minimum(2/3 ,-1/3)

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extreme f(x)=x^3-8y^3+4xy-8

extreme

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 48 y

D (x, y) = - 288 x y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

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

What is the extreme f(x)=x^3-8y^3+4xy-8 ?
The extreme f(x)=x^3-8y^3+4xy-8 is Saddle(0,0),Minimum(2/3 ,-1/3)