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

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

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extreme f(x,y)=x^2+2y^2-2x-4y+1

extreme points

f (x, y) = x^{2} + 2 y^{2} - 2 x - 4 y + 1

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x, y) = 2 x^{2} + y^{2} + 8 x - 6 y - 2 x y + 12

f (x) = x^{3} - 3 x^{2} + 24 x

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

f (x) = \frac{x ^{2}}{x + 4}

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

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