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)

y = \frac{I n x}{x}

f (x) = lo g_{2} (\frac{2 - 3 x}{3 - x})

f (x, y) = - x^{2} - 5 y^{2} + 8 x - 10 y

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

f (x) = 3 x^{2} - 150 ln (x)

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)