What is the extreme f(x,y)=sqrt(400-49x^2-16y^2) ?

The extreme f(x,y)=sqrt(400-49x^2-16y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-49x^2-16y^2)

extreme points

f (x, y) = 400 - 49 x^{2} - 16 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{16 ( 400 - 49 x ^{2} )}{( 400 - 49 x ^{2} - 16 y ^{2} ) 400 - 49 x ^{2} - 16 y ^{2}}

D (x, y) = \frac{313600}{( 400 - 49 x ^{2} - 16 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{3 x}{x ^{2} + 1}, - 4 \leq x \leq 4

f (x) = x^{3} + y^{3} - 3 x^{2} - 6 y^{2} - 9 x

2 x^{3} - x^{2} - 4 x + 12

f (x) = x^{3} - \frac{3}{2} x^{2} - 6 x + 1

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

What is the extreme f(x,y)=sqrt(400-49x^2-16y^2) ?
The extreme f(x,y)=sqrt(400-49x^2-16y^2) is Maximum(0,0)