What is the extreme f(x,y)=sqrt(400-9x^2-25y^2) ?

The extreme f(x,y)=sqrt(400-9x^2-25y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-9x^2-25y^2)

extreme points

f (x, y) = 400 - 9 x^{2} - 25 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{25 ( 400 - 9 x ^{2} )}{( 400 - 9 x ^{2} - 25 y ^{2} ) 400 - 9 x ^{2} - 25 y ^{2}}

D (x, y) = \frac{90000}{( 400 - 9 x ^{2} - 25 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{e ^{x}}{x ^{5}}, x > 0

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

f (x) = 4 x^{2} - 16 x + 4, 0 \leq x \leq 3

f (x, y) = 2 x^{2} + 4 x + 2 y^{2} + 5 y

f (x) = x^{3} - 2^{2}

What is the extreme f(x,y)=sqrt(400-9x^2-25y^2) ?
The extreme f(x,y)=sqrt(400-9x^2-25y^2) is Maximum(0,0)