What is the extreme f(x,y)=-3xy+x^3-y^3+1/8 ?

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

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extreme f(x,y)=-3xy+x^3-y^3+1/8

extreme points

f (x, y) = - 3 x y + x^{3} - y^{3} + \frac{1}{8}

Saddle (0, 0), Maximum (- 1, 1)

Solution steps

Find the critical points:

(0, 0), (- 1, 1)

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

D (x, y) = - 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 1) :

Maximum

Saddle (0, 0), Maximum (- 1, 1)

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