What is the extreme P(x,y)=16x^2-9y^2 ?

The extreme P(x,y)=16x^2-9y^2 is Saddle(0,0)

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extreme P(x,y)=16x^2-9y^2

extreme points

P (x, y) = 16 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) = - 576

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 1920 x - 10 x^{3}

f (x) = 100 e^{- 0.062 \cdot x}

f (x) = 7 sin^{2} (x) - 14 cos (x)

f (x) = - 3 x^{2} - 30 x - 21

y = z - x

What is the extreme P(x,y)=16x^2-9y^2 ?
The extreme P(x,y)=16x^2-9y^2 is Saddle(0,0)