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 points

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

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)

x + 3, 0 \leq x \leq 2

\frac{x}{x ^{2} - x + 1} (0.3)

f (x) = x^{2} + \frac{200}{x}, (0, \infty)

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

f (x) = 5 x - 8 sin (x), 0 \leq x \leq π

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)