What is the extreme f(x,y)=x^3+y^3-9xy+27 ?

The extreme f(x,y)=x^3+y^3-9xy+27 is Saddle(0,0),Minimum(3,3)

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extreme f(x,y)=x^3+y^3-9xy+27

extreme points

f (x, y) = x^{3} + y^{3} - 9 x y + 27

Saddle (0, 0), Minimum (3, 3)

Solution steps

Find the critical points:

(0, 0), (3, 3)

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

D (x, y) = 36 x y - 81

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(3, 3) :

Minimum

Saddle (0, 0), Minimum (3, 3)

f (x) = 6 x^{2} ln (x)

x^{4} - 4 x^{3} + 1

f (x, y) = \frac{x ^{4} - y ^{4}}{x ^{2} - y ^{2}}

y = - 2 x^{3} + 6 x^{2} - 3

f (x) = x ln (x^{2} - y)

What is the extreme f(x,y)=x^3+y^3-9xy+27 ?
The extreme f(x,y)=x^3+y^3-9xy+27 is Saddle(0,0),Minimum(3,3)