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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = (x^{2} - 5)^{4} (x^{2} + 1)^{5}

f (x) = 240 x + 57 x^{2} - x^{3}

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

f (x) = x^{2} + 2 x + 5

\frac{x ^{2} - 12}{x - 4}

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