What is the extreme f(x,y)=y^3-x^3-2xy+6 ?

The extreme f(x,y)=y^3-x^3-2xy+6 is Saddle(0,0),Maximum(2/3 ,-2/3)

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extreme f(x,y)=y^3-x^3-2xy+6

extreme points

f (x, y) = y^{3} - x^{3} - 2 x y + 6

Saddle (0, 0), Maximum (\frac{2}{3}, - \frac{2}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{2}{3}, - \frac{2}{3})

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

D (x, y) = - 36 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{2}{3}, - \frac{2}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{2}{3}, - \frac{2}{3})

f (x) = x^{3} - 3 x^{2} - 9 x + 12

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, (0, 1)

f (x, y) = x^{2} + y^{2} + 12 x - 16 y

f (x, y) = \frac{3}{x ^{2} + y ^{2} - 9}

6 x + 12

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