What is the extreme f(x,y)=sqrt(400-25x^2-64y^2) ?

The extreme f(x,y)=sqrt(400-25x^2-64y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-25x^2-64y^2)

extreme points

f (x, y) = 400 - 25 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 - 25 x ^{2} )}{( 400 - 25 x ^{2} - 64 y ^{2} ) 400 - 25 x ^{2} - 64 y ^{2}}

D (x, y) = \frac{640000}{( 400 - 25 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) = x^{3} + 3 x^{2} - 9 x - 1

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

f (x) = x^{2} - 8 x + 9

f (x, y) = x^{3} - y^{3} + 6 x y

f (x, y) = 288 y^{2} + x^{2} - x^{2} y

What is the extreme f(x,y)=sqrt(400-25x^2-64y^2) ?
The extreme f(x,y)=sqrt(400-25x^2-64y^2) is Maximum(0,0)