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, y) = \frac{500}{4 + x ^{2} + y ^{2}}

f (x) = x^{3} + 5 x - 4

f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

f (x) = x 5 - x

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