What is the extreme f(x,y)=sqrt(400-100x^2-16y^2) ?

The extreme f(x,y)=sqrt(400-100x^2-16y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-100x^2-16y^2)

extreme points

f (x, y) = 400 - 100 x^{2} - 16 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{8 ( 100 - 25 x ^{2} )}{( 100 - 25 x ^{2} - 4 y ^{2} ) 100 - 25 x ^{2} - 4 y ^{2}}

D (x, y) = \frac{40000}{( 100 - 25 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, y) = x^{3} + y^{3} + 3 x y^{2} - 18 (x + y)

x, f = \frac{20000}{3 + x ^{2} + y ^{2}}

f (x) = - (x^{2} - 4)

y = - 4 e^{- 4 t} - 5 e^{- 5 t}

f (x) = x^{4} - 24 x^{2} + 23

What is the extreme f(x,y)=sqrt(400-100x^2-16y^2) ?
The extreme f(x,y)=sqrt(400-100x^2-16y^2) is Maximum(0,0)