What is the extreme P(x,y)=10x^2-10y^2 ?

The extreme P(x,y)=10x^2-10y^2 is Saddle(0,0)

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extreme P(x,y)=10x^2-10y^2

extreme points

P (x, y) = 10 x^{2} - 10 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 20

D (x, y) = - 400

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{2} e^{1 - x}

(3 - x) (x - 5)^{2}

f (x, y) = \frac{1}{x} + \frac{1}{y} - x y + 3

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

f (x) = \frac{( x )}{( 6 x + 2 )}, 5 \leq x \leq 8

What is the extreme P(x,y)=10x^2-10y^2 ?
The extreme P(x,y)=10x^2-10y^2 is Saddle(0,0)