What is the extreme f(x,y)=9x^2+18xy+y^3 ?

The extreme f(x,y)=9x^2+18xy+y^3 is Saddle(0,0),Minimum(-6,6)

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extreme f(x,y)=9x^2+18xy+y^3

extreme points

f (x, y) = 9 x^{2} + 18 x y + y^{3}

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) = 108 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) = 2 x^{2} + 3 x - 5

f (x) = x^{3} - 12 x - 10

g (x) = x^{2} - 6 x - 12

f (x) = - 6 x + 5 y

f (x) = - \frac{1}{2} x^{4} + 2 x^{3} + 40 x^{2} - 13

What is the extreme f(x,y)=9x^2+18xy+y^3 ?
The extreme f(x,y)=9x^2+18xy+y^3 is Saddle(0,0),Minimum(-6,6)