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

The extreme f(x)=3x^3+y^3-9xy+27 is Saddle(0,0),Minimum(\sqrt[3]{3},3^{2/3})

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

extreme points

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

Saddle (0, 0), Minimum (33, 3^{\frac{2}{3}})

Solution steps

Find the critical points:

(0, 0), (33, 3^{\frac{2}{3}})

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

D (x, y) = 108 x y - 81

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(33, 3^{\frac{2}{3}}) :

Minimum

Saddle (0, 0), Minimum (33, 3^{\frac{2}{3}})

f (x) = 15 + 398 x + 184 y - 9 x^{2} - 2 y^{2} - 8 x y

f (x) = cos (x), - \frac{π}{4} \leq x \leq \frac{π}{4}

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

f (x) = x y + \frac{8}{x}

f (x, y) = x^{(2)} - 2 x y + 4 y^{(2)} - 4 x - 2 y + 24

What is the extreme f(x)=3x^3+y^3-9xy+27 ?
The extreme f(x)=3x^3+y^3-9xy+27 is Saddle(0,0),Minimum(\sqrt[3]{3},3^{2/3})