What is the extreme f(x,y)=sqrt(48-3x^2-3y^2) ?

The extreme f(x,y)=sqrt(48-3x^2-3y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(48-3x^2-3y^2)

extreme points

f (x, y) = 48 - 3 x^{2} - 3 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{3 ( 16 - x ^{2} )}{( 16 - x ^{2} - y ^{2} ) 48 - 3 x ^{2} - 3 y ^{2}}

D (x, y) = \frac{48}{( 16 - x ^{2} - y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x) = x e^{6 x^{2}}

f (x) = 2 x^{3} - 6 x + 4

f (x, y) = x^{2} y + y^{3} - 75 y

f (x, y) = e^{- x - y}

What is the extreme f(x,y)=sqrt(48-3x^2-3y^2) ?
The extreme f(x,y)=sqrt(48-3x^2-3y^2) is Maximum(0,0)