What is the extreme f(x)=8x-2x^2+12xy-12y^2+y ?

The extreme f(x)=8x-2x^2+12xy-12y^2+y is Saddle(-17/4 ,-25/12)

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extreme f(x)=8x-2x^2+12xy-12y^2+y

extreme points

f (x) = 8 x - 2 x^{2} + 12 x y - 12 y^{2} + y

Saddle (- \frac{17}{4}, - \frac{25}{12})

Solution steps

Find the critical points:

(- \frac{17}{4}, - \frac{25}{12})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 24

D (x, y) = - 48

Check the sign of at each critical point

Check critical point

(- \frac{17}{4}, - \frac{25}{12}) :

Saddle

Saddle (- \frac{17}{4}, - \frac{25}{12})

f (x) = - 3 x^{3} + 9 x^{2} - 8

f (x) = - 3 x^{3} + 9 x^{2} - 4

f (x, y) = x^{2} + y^{2} + 2 x - 6 y - 3

f (x y) = \frac{3 y + 5 x}{x y + 3}

f (x, y) = - 4 x^{2} + 8 x y - 2 y^{4} - 4

What is the extreme f(x)=8x-2x^2+12xy-12y^2+y ?
The extreme f(x)=8x-2x^2+12xy-12y^2+y is Saddle(-17/4 ,-25/12)