What is the extreme f(x,y)=x^3-y^3-2xy+10 ?

The extreme f(x,y)=x^3-y^3-2xy+10 is Saddle(0,0),Maximum(-2/3 , 2/3)

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extreme f(x,y)=x^3-y^3-2xy+10

extreme points

f (x, y) = x^{3} - y^{3} - 2 x y + 10

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})

y = x^{4} - 18 x^{2}

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

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (0, 10)

x^{2} y + x y^{2} + 3 x y

\frac{x ^{2} + 4 x + 3}{x ^{3} - 2 x ^{2} - 5 x + 6}

What is the extreme f(x,y)=x^3-y^3-2xy+10 ?
The extreme f(x,y)=x^3-y^3-2xy+10 is Saddle(0,0),Maximum(-2/3 , 2/3)