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

The extreme f(x,y)=3x^2-4xy+4y^2-4x+8y+4 is Minimum(0,-1)

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extreme f(x,y)=3x^2-4xy+4y^2-4x+8y+4

extreme points

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

Minimum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 32

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Minimum (0, - 1)

3 - x

f (x) = 2 x^{3} - 18 x^{2} + 48 x - 7

x^{3} + y^{3} - 27 x - 48 y - 24

f (x) = e^{- x} \cdot (x^{2} - x + 1)

f (x) = - 2 x^{2} + 60 x - 70

What is the extreme f(x,y)=3x^2-4xy+4y^2-4x+8y+4 ?
The extreme f(x,y)=3x^2-4xy+4y^2-4x+8y+4 is Minimum(0,-1)