What is the extreme f(x,y)=sqrt(400-64x^2-36y^2) ?

The extreme f(x,y)=sqrt(400-64x^2-36y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-64x^2-36y^2)

extreme points

f (x, y) = 400 - 64 x^{2} - 36 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{18 ( 100 - 16 x ^{2} )}{( 100 - 16 x ^{2} - 9 y ^{2} ) 100 - 16 x ^{2} - 9 y ^{2}}

D (x, y) = \frac{57600}{( 100 - 16 x ^{2} - 9 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

x e^{\frac{( 5 - x )}{4}}

- x^{4} + 3 x^{3} - x^{2} - 5 x + 7

f (x) = - 3 x^{3} - 9

f (x) = - \frac{1}{2} x^{2} + 5 - 8

f (x) = - \frac{1}{10} x^{3} + 9 x^{2} + 500

What is the extreme f(x,y)=sqrt(400-64x^2-36y^2) ?
The extreme f(x,y)=sqrt(400-64x^2-36y^2) is Maximum(0,0)