What is the extreme f(x,y)=sqrt(400-49x^2-81v^2) ?

The extreme f(x,y)=sqrt(400-49x^2-81v^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-49x^2-81v^2)

extreme points

f (x, y) = 400 - 49 x^{2} - 81 v^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{81 ( 400 - 49 x ^{2} )}{( 400 - 49 x ^{2} - 81 v ^{2} ) 400 - 49 x ^{2} - 81 v ^{2}}

D (x, v) = \frac{1587600}{( 400 - 49 x ^{2} - 81 v ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = 3 x y - 2 x^{2} - 2 y^{2} + 28 x - 14 y - 3

f (x) = 2 x^{2} e^{- \frac{x}{5}}

f (x, y) = 3 x y + x^{3} - y^{3} + 1

f (x) = 9 - 2 x + 4 y - x^{2} - 2 y^{3}

y = x (6 - 2 x)^{2}

What is the extreme f(x,y)=sqrt(400-49x^2-81v^2) ?
The extreme f(x,y)=sqrt(400-49x^2-81v^2) is Maximum(0,0)