What is the extreme f(x,y)=x^3-y^3-6xy-4 ?

The extreme f(x,y)=x^3-y^3-6xy-4 is Saddle(0,0),Maximum(-2,2)

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extreme f(x,y)=x^3-y^3-6xy-4

extreme points

f (x, y) = x^{3} - y^{3} - 6 x y - 4

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) = \frac{2 ( x + 2 ) ^{2}}{x ^{2}}

f (x, y) = y^{3} x^{5} + y x

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

2 x^{3} - 3 x^{2} - 12 x + 8

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

What is the extreme f(x,y)=x^3-y^3-6xy-4 ?
The extreme f(x,y)=x^3-y^3-6xy-4 is Saddle(0,0),Maximum(-2,2)