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

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

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

extreme

Saddle (- \frac{1}{3}, - \frac{1}{3}), Minimum (\frac{1}{2}, - \frac{3}{4})

Solution steps

Find the critical points:

(- \frac{1}{3}, - \frac{1}{3}), (\frac{1}{2}, - \frac{3}{4})

\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

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

Saddle

Check critical point

(\frac{1}{2}, - \frac{3}{4}) :

Minimum

Saddle (- \frac{1}{3}, - \frac{1}{3}), Minimum (\frac{1}{2}, - \frac{3}{4})

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