What is the extreme f(x,y)=x^2+2y^2-4x+4y+6 ?

The extreme f(x,y)=x^2+2y^2-4x+4y+6 is Minimum(2,-1)

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extreme f(x,y)=x^2+2y^2-4x+4y+6

extreme points

f (x, y) = x^{2} + 2 y^{2} - 4 x + 4 y + 6

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

f (x) = ln (4 x) - 8 x^{2}

3 x^{4} + 3 y^{4} - 2 x y

f (x) = 150 x + 18 x^{2} - 1.5 x^{3}

f (x) = sin^{2} (2 x) - cos^{2} (2 x) - 1

f (x) = 2 x - 4 cos (x), - 2 \leq x \leq 0

What is the extreme f(x,y)=x^2+2y^2-4x+4y+6 ?
The extreme f(x,y)=x^2+2y^2-4x+4y+6 is Minimum(2,-1)