What is the extreme f(x,y)=sqrt(400-49x^2-100y^2) ?

The extreme f(x,y)=sqrt(400-49x^2-100y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-49x^2-100y^2)

extreme points

f (x, y) = 400 - 49 x^{2} - 100 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{100 ( 400 - 49 x ^{2} )}{( 400 - 49 x ^{2} - 100 y ^{2} ) 400 - 49 x ^{2} - 100 y ^{2}}

D (x, y) = \frac{1960000}{( 400 - 49 x ^{2} - 100 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 3 x^{4} - 8 x^{3} - 144 x^{2} + 1

f (x, y) = - 5 x^{2} - 5 x y - 7 y^{2} + 50 x - 67 y + 9

y = x^{2} + 2 x + 2

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

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

What is the extreme f(x,y)=sqrt(400-49x^2-100y^2) ?
The extreme f(x,y)=sqrt(400-49x^2-100y^2) is Maximum(0,0)