What is the extreme f(x,y)=sqrt(x^2+y^2+64) ?

The extreme f(x,y)=sqrt(x^2+y^2+64) is Minimum(0,0)

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extreme f(x,y)=sqrt(x^2+y^2+64)

extreme points

f (x, y) = x^{2} + y^{2} + 64

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 64}{( x ^{2} + y ^{2} + 64 ) x ^{2} + y ^{2} + 64}

D (x, y) = \frac{64}{( x ^{2} + y ^{2} + 64 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

x^{2} + 2

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

f (x, y) = 40 x^{\frac{3}{4}} y^{\frac{1}{4}}

y = 9 e^{- 2 t} - 7 e^{- 3 t}

18 x - 8

What is the extreme f(x,y)=sqrt(x^2+y^2+64) ?
The extreme f(x,y)=sqrt(x^2+y^2+64) is Minimum(0,0)