What is the extreme f(x,y)=x^3+6xy+y^3 ?

The extreme f(x,y)=x^3+6xy+y^3 is Saddle(0,0),Maximum(-2,-2)

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extreme f(x,y)=x^3+6xy+y^3

extreme points

f (x, y) = x^{3} + 6 x y + y^{3}

Saddle (0, 0), Maximum (- 2, - 2)

Solution steps

Find the critical points:

(0, 0), (- 2, - 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, - 2) :

Maximum

Saddle (0, 0), Maximum (- 2, - 2)

f (x, y) = x y - 5 x + 15

f (x, y) = ln (4 x^{2} + 9 y^{2} + 36)

f (x) = \frac{x}{x ^{2} + 5 x + 4}

2 ln (1 + x^{2})

(x - 1) e^{x + 1}

What is the extreme f(x,y)=x^3+6xy+y^3 ?
The extreme f(x,y)=x^3+6xy+y^3 is Saddle(0,0),Maximum(-2,-2)