What is the extreme f(x,y)=-x^3+y^3+15xy+1 ?

The extreme f(x,y)=-x^3+y^3+15xy+1 is Saddle(0,0),Minimum(-5,5)

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extreme f(x,y)=-x^3+y^3+15xy+1

extreme

Saddle (0, 0), Minimum (- 5, 5)

Solution steps

Find the critical points:

(0, 0), (- 5, 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = - 36 x y - 225

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 5, 5) :

Minimum

Saddle (0, 0), Minimum (- 5, 5)

What is the extreme f(x,y)=-x^3+y^3+15xy+1 ?
The extreme f(x,y)=-x^3+y^3+15xy+1 is Saddle(0,0),Minimum(-5,5)