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

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

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

extreme points

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

Saddle (0, 0), Minimum (9, 9)

Solution steps

Find the critical points:

(0, 0), (9, 9)

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

D (x, y) = 36 x y - 729

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(9, 9) :

Minimum

Saddle (0, 0), Minimum (9, 9)

f (x) = y^{5} - 3 x y

f (x, y) = x^{2} + y^{2} - 2 x + 5 y

f (x) = 3 - 4 x - \frac{4}{x}

f (x, y) = - 3 x^{2} - 4 x y - 8 y^{2} + 42 x + 68 y + 4

f (x) = \frac{1}{3} x^{3} - x^{2} + 3 x + 4

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