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

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

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

extreme

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)

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