What is the extreme f(x,y)=sqrt(400-4x^2-81y^2) ?

The extreme f(x,y)=sqrt(400-4x^2-81y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-4x^2-81y^2)

extreme points

f (x, y) = 400 - 4 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 - 4 x ^{2} )}{( 400 - 4 x ^{2} - 81 y ^{2} ) 400 - 4 x ^{2} - 81 y ^{2}}

D (x, y) = \frac{129600}{( 400 - 4 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) = \frac{( x + 5 )}{x ^{2} - 2 x - 35}

f (x) = 2 x - 5 y + 3

f (x) = x^{2} + 2 y^{2} + 3 x + 10

f (x) = 8 cos (θ) + 7 sin (θ)

(11 - x) (x + 1)^{2}

What is the extreme f(x,y)=sqrt(400-4x^2-81y^2) ?
The extreme f(x,y)=sqrt(400-4x^2-81y^2) is Maximum(0,0)