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

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

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

extreme points

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

Minimum (0, 2)

Solution steps

Find the critical points:

(0, 2)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Minimum (0, 2)

f (x) = 2 - 5 x + x^{2}

f (x, y) = 33 x + 66 y + x y - x^{2} - 3 y^{2}

f (x) = 3 x + 4 32.5 - x^{2}

8 x^{2} - 8 x^{4}

f (x) = 6 x^{2} - 10 y^{2}

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