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) = x^{2} + 2 x y + 2 y^{2} + 2 x - 2 y

f (x) = x^{3} - \frac{3}{2} x^{2}, - 3 \leq x \leq 6

f (x, y) = e^{2 x} - x^{3} y^{5}

f (x) = 3 x^{4} + x - 1

x + \frac{108}{x ^{2}}

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)