What is the extreme f(x,y)=sqrt(400-16x^2-81y^2) ?

The extreme f(x,y)=sqrt(400-16x^2-81y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-16x^2-81y^2)

extreme points

f (x, y) = 400 - 16 x^{2} - 81 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{81 ( 400 - 16 x ^{2} )}{( 400 - 16 x ^{2} - 81 y ^{2} ) 400 - 16 x ^{2} - 81 y ^{2}}

D (x, y) = \frac{518400}{( 400 - 16 x ^{2} - 81 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = ln (7 + x^{2})

y = - x^{3} + \frac{3}{2} x^{2} + 18 x

3 x + 3

(x - 2)^{3} + 1

\frac{5 x}{x ^{2} + 9}

What is the extreme f(x,y)=sqrt(400-16x^2-81y^2) ?
The extreme f(x,y)=sqrt(400-16x^2-81y^2) is Maximum(0,0)