What is the extreme f(x,y)=x^3+y^3-18xy ?

The extreme f(x,y)=x^3+y^3-18xy is Saddle(0,0),Minimum(6,6)

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extreme f(x,y)=x^3+y^3-18xy

extreme points

f (x, y) = x^{3} + y^{3} - 18 x y

Saddle (0, 0), Minimum (6, 6)

Solution steps

Find the critical points:

(0, 0), (6, 6)

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

D (x, y) = 36 x y - 324

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(6, 6) :

Minimum

Saddle (0, 0), Minimum (6, 6)

f (x) = x^{2} + \frac{8}{x}

f (t) = 2 t + 3 e^{- k t}

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

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

f (x, y) = 7 + 2 x + 4 y - x^{2} - 4 y^{2}

What is the extreme f(x,y)=x^3+y^3-18xy ?
The extreme f(x,y)=x^3+y^3-18xy is Saddle(0,0),Minimum(6,6)