What is the extreme f(x,y)=x^2+y^2-8y+16 ?

The extreme f(x,y)=x^2+y^2-8y+16 is Minimum(0,4)

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extreme f(x,y)=x^2+y^2-8y+16

extreme points

f (x, y) = x^{2} + y^{2} - 8 y + 16

Minimum (0, 4)

Solution steps

Find the critical points:

(0, 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 4) :

Minimum

Minimum (0, 4)

\frac{3 x - 6}{x + 2}

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

f (x) = 4 sin (x) + 4 cos (x)

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

f (x) = \frac{6}{x + 7}

What is the extreme f(x,y)=x^2+y^2-8y+16 ?
The extreme f(x,y)=x^2+y^2-8y+16 is Minimum(0,4)