What is the extreme f(x,y)=18x^2-32y^2-36x-128y-110 ?

The extreme f(x,y)=18x^2-32y^2-36x-128y-110 is Saddle(1,-2)

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extreme f(x,y)=18x^2-32y^2-36x-128y-110

extreme points

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

Saddle (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 64

D (x, y) = - 2304

Check the sign of at each critical point

Check critical point

(1, - 2) :

Saddle

Saddle (1, - 2)

f (x, y) = x^{3} + y^{3} - 24 x y

f (x) = \frac{8}{x} + 2 π x^{2}

f (x, y) = 400 - 9 x^{2} - 36 y^{2}

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

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

What is the extreme f(x,y)=18x^2-32y^2-36x-128y-110 ?
The extreme f(x,y)=18x^2-32y^2-36x-128y-110 is Saddle(1,-2)