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

The extreme f(x,y)=x^3+xy+y^2 is Saddle(0,0),Minimum(1/6 ,-1/12)

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

extreme points

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

Saddle (0, 0), Minimum (\frac{1}{6}, - \frac{1}{12})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{6}, - \frac{1}{12})

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

D (x, y) = 12 x - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{6}, - \frac{1}{12}) :

Minimum

Saddle (0, 0), Minimum (\frac{1}{6}, - \frac{1}{12})

What is the extreme f(x,y)=x^3+xy+y^2 ?
The extreme f(x,y)=x^3+xy+y^2 is Saddle(0,0),Minimum(1/6 ,-1/12)