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

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

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

extreme points

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

Minimum (1, - 1)

Solution steps

Find the critical points:

(1, - 1)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Minimum (1, - 1)

2 x^{2} + 8 x + 12

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

(x - 9) ln (x - 9)

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

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

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