What is the extreme f(x,y)=sqrt(400-16x^2-25y^2) ?

The extreme f(x,y)=sqrt(400-16x^2-25y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-16x^2-25y^2)

extreme points

f (x, y) = 400 - 16 x^{2} - 25 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{25 ( 400 - 16 x ^{2} )}{( 400 - 16 x ^{2} - 25 y ^{2} ) 400 - 16 x ^{2} - 25 y ^{2}}

D (x, y) = \frac{160000}{( 400 - 16 x ^{2} - 25 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = sin (11 x)

f (x, y) = \frac{1}{4} x^{4} - \frac{1}{2} y^{2} - \frac{1}{2} x^{2} - y + 10

f (x) = x^{3} + y^{3} - 27 x - 48 y - 24

f (x) = 4 (5 x y + \frac{1024}{y} + \frac{1024}{x})

f (x) = - 100 x^{2} + 1800 x

What is the extreme f(x,y)=sqrt(400-16x^2-25y^2) ?
The extreme f(x,y)=sqrt(400-16x^2-25y^2) is Maximum(0,0)