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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x^{2}

D (x, y) = - 48 x^{2} y^{2} + 48 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = x + \frac{5}{x}

y = x 36 - x^{2}

x^{2} - 8 ln (x)

f (x, y) = x y^{2} - 6 x^{2} - 3 x + 2 x y + 9

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