What is the extreme P(x,y)=3x^2-8xy-3y^2 ?

The extreme P(x,y)=3x^2-8xy-3y^2 is Saddle(0,0)

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extreme P(x,y)=3x^2-8xy-3y^2

extreme points

P (x, y) = 3 x^{2} - 8 x y - 3 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = - 100

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{4} - 2 x^{2} + 8

f (x) = - \frac{1}{x - 2}

f (x) = (8 - (3 x^{2} + 2 x + 1))

f (x) = \frac{( 2 x ^{2} )}{9 - x ^{2}}

f (x) = - \frac{1}{x - 6}

What is the extreme P(x,y)=3x^2-8xy-3y^2 ?
The extreme P(x,y)=3x^2-8xy-3y^2 is Saddle(0,0)