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

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

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

extreme points

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

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}} = - 2

D (x, y) = 12 x - 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) = 3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x

x^{3} - 3 x^{2} - 9 x + 1

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

f (x) = x^{2} - y^{2} - 2 x + 4 y + 6

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6, 0 \leq x \leq 3

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