What is the extreme f(x,y)=sqrt(2x^2+2y^2-4y+9) ?

The extreme f(x,y)=sqrt(2x^2+2y^2-4y+9) is Minimum(0,1)

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extreme f(x,y)=sqrt(2x^2+2y^2-4y+9)

extreme points

f (x, y) = 2 x^{2} + 2 y^{2} - 4 y + 9

Minimum (0, 1)

Solution steps

Find the critical points:

(0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( 2 x ^{2} + 7 )}{( 2 x ^{2} + 2 y ^{2} - 4 y + 9 ) 2 x ^{2} + 2 y ^{2} - 4 y + 9}

D (x, y) = \frac{28}{( 9 + 2 x ^{2} + 2 y ^{2} - 4 y ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Minimum (0, 1)

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

f (x) = x (x^{3} - x + 10)

f (x) = - 462 x + 64 x^{2} - 2 x^{3}

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

f (x, y) = 3 x^{2} + 2 y^{2} + 6 x + 8 y

What is the extreme f(x,y)=sqrt(2x^2+2y^2-4y+9) ?
The extreme f(x,y)=sqrt(2x^2+2y^2-4y+9) is Minimum(0,1)