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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 18 x^{3} y

D (x, y) = - 405 x^{4} y^{4} + 108 x^{2} y^{2} - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - x^{3} - 9 x^{2} + 165 x + 1300

f (x, y) = (x^{2} - 1) (y^{2} - 1)

\frac{2}{3} x^{3} - 4 x^{3} + 12

f (x) = \frac{4860}{x} + 15 x + 700508

2 sec (t) + tan (t)

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