What is the extreme f(x,y)=Z=x^2-3xy+2y^2-4y ?

The extreme f(x,y)=Z=x^2-3xy+2y^2-4y is Saddle(-12,-8)

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extreme f(x,y)=Z=x^2-3xy+2y^2-4y

extreme points

f (x, y) = Z = x^{2} - 3 x y + 2 y^{2} - 4 y

Saddle (- 12, - 8)

Solution steps

Find the critical points:

(- 12, - 8)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(- 12, - 8) :

Saddle

Saddle (- 12, - 8)

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 3 y + 10

2 t^{4} + 6 t^{3} - 5

f (x, y) = arcsec (x^{2} - ∣ y ∣)

f (x) = x^{\frac{2}{5}}

f (x) = 2 x^{2} - 4 x y + 3 y^{2} - 8 x + 8 y - 1 = 0

What is the extreme f(x,y)=Z=x^2-3xy+2y^2-4y ?
The extreme f(x,y)=Z=x^2-3xy+2y^2-4y is Saddle(-12,-8)