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

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

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

extreme points

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

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, y) = 2 x^{2} + 3 y^{2} - 12 x - 2 y + 1

x^{2} (x - 1) (2 x + 1) + y^{2}

f (x, y) = \frac{x y}{x ^{2} + y ^{2} - 25}

f (x) = x^{4} + 4 x^{2} - 10

f (x) = - x^{2} + 6

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