What is the extreme f(x,y)=sqrt(400-81x^2-4y^2) ?

The extreme f(x,y)=sqrt(400-81x^2-4y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-81x^2-4y^2)

extreme points

f (x, y) = 400 - 81 x^{2} - 4 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 ( 400 - 81 x ^{2} )}{( 400 - 81 x ^{2} - 4 y ^{2} ) 400 - 81 x ^{2} - 4 y ^{2}}

D (x, y) = \frac{129600}{( 400 - 81 x ^{2} - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 9 t^{2} + 4 t - 10

(x^{2} - 4)^{\frac{1}{7}}, - 2 \leq x \leq 3

f (x) = \frac{3}{2} t^{4} - t^{6}

f (x, y) = - x^{2} - y^{3} + 3 y^{2} - 2 x + 45 y + 1

y = - 12 x^{5} - 45 x^{4} + 20 x^{3} - 1

What is the extreme f(x,y)=sqrt(400-81x^2-4y^2) ?
The extreme f(x,y)=sqrt(400-81x^2-4y^2) is Maximum(0,0)