What is the extreme f(x,y)=sqrt(50-2x^2-2y^2) ?

The extreme f(x,y)=sqrt(50-2x^2-2y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(50-2x^2-2y^2)

extreme points

f (x, y) = 50 - 2 x^{2} - 2 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 ( 25 - x ^{2} )}{( 25 - x ^{2} - y ^{2} ) 50 - 2 x ^{2} - 2 y ^{2}}

D (x, y) = \frac{50}{( 25 - x ^{2} - y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (p, q) = p^{2} - 2 p + 0.8 q^{2}

y = 2 + 12 x - x^{3}

y = \frac{x ^{3}}{3} - 2 x^{2} - 5 x

f (x) = x^{3} - 3 x^{2} + 12, - 2 \leq x \leq 4

f (x) = \frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x

What is the extreme f(x,y)=sqrt(50-2x^2-2y^2) ?
The extreme f(x,y)=sqrt(50-2x^2-2y^2) is Maximum(0,0)