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})

f (x) = 2 x^{2} - 8 x, 0 \leq x \leq 6

x^{2} + 2 x - 8

y = 2 x^{3} - 21 x^{2} + 60 x + 4

f (x) = (x + 1)^{5} - 5 x - 3

a^{3} y = x^{2} (2 a^{2} x^{2})

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)