What is the extreme P(X,Y)=36X^2-9Y^2 ?

The extreme P(X,Y)=36X^2-9Y^2 is Saddle(0,0)

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extreme P(X,Y)=36X^2-9Y^2

extreme points

P (X, Y) = 36 X^{2} - 9 Y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial X ^{2}} = - 18

D (X, Y) = - 1296

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

\frac{4}{x}

f (x) = - 4 x^{2} + 40 x - 60

\frac{1}{x ^{2} + 2 x - 8}

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

What is the extreme P(X,Y)=36X^2-9Y^2 ?
The extreme P(X,Y)=36X^2-9Y^2 is Saddle(0,0)