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

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

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

extreme points

f (x) = 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)

- 45 x^{2} - 27 x + 2

y = x^{3} - 9 x^{2} - 48 x

f (x, y) = 100 (y - x^{2})^{2} + (1 - x)^{2}

f (x) = - (5 x + 3) e^{- 2 x}

\frac{x ^{2}}{x ^{2} + 2 x - 15}

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