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

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

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

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 16}{( x ^{2} + y ^{2} + 16 ) x ^{2} + y ^{2} + 16}

D (x, y) = \frac{16}{( x ^{2} + y ^{2} + 16 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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