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

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

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

extreme points

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

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)

f (x) = x^{3} - 3 x^{2} - 9 x + 10

f (x) = x 4 - x^{2}, - 1 \leq x \leq 2

f (x) = \frac{x ^{2}}{x ^{2} - 3}

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

f (x) = \frac{1}{x + 2}

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