What is the extreme f(x,y)=sqrt(100-x^2-y^2) ?

The extreme f(x,y)=sqrt(100-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(100-x^2-y^2)

extreme points

f (x, y) = 100 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{100 - x ^{2}}{( 100 - x ^{2} - y ^{2} ) 100 - x ^{2} - y ^{2}}

D (x, y) = \frac{100}{( - x ^{2} - y ^{2} + 100 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

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

f (x) = 2 sin (x) - cos (2 x)

y = x^{3} - 3 x

f (x) = e^{2 x} + e^{- x}

What is the extreme f(x,y)=sqrt(100-x^2-y^2) ?
The extreme f(x,y)=sqrt(100-x^2-y^2) is Maximum(0,0)