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)

\frac{5}{- 5 e x ^{2} - 5 y ^{2} + 5}

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

f (x) = 6 + x^{2} + 4 x + y^{2}

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

f (x, y) = x y + \frac{50}{x} + \frac{20}{y}

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