What is the extreme f(x,y)=sqrt(400-49x^2-64y^2) ?

The extreme f(x,y)=sqrt(400-49x^2-64y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-49x^2-64y^2)

extreme points

f (x, y) = 400 - 49 x^{2} - 64 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{64 ( 400 - 49 x ^{2} )}{( 400 - 49 x ^{2} - 64 y ^{2} ) 400 - 49 x ^{2} - 64 y ^{2}}

D (x, y) = \frac{1254400}{( 400 - 49 x ^{2} - 64 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x ln ∣ x ∣

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

f (x) = x^{3} + 8

f (x) = \frac{900}{x} + 2 π x^{2}

\frac{ln ( x )}{x ^{3}}

What is the extreme f(x,y)=sqrt(400-49x^2-64y^2) ?
The extreme f(x,y)=sqrt(400-49x^2-64y^2) is Maximum(0,0)