What is the extreme f(x,y)=sqrt(400-16x^2-64y^2) ?

The extreme f(x,y)=sqrt(400-16x^2-64y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(400-16x^2-64y^2)

extreme points

f (x, y) = 400 - 16 x^{2} - 64 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{16 ( 25 - x ^{2} )}{( 25 - x ^{2} - 4 y ^{2} ) 25 - x ^{2} - 4 y ^{2}}

D (x, y) = \frac{1600}{( - x ^{2} + 25 - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = y^{3} + x^{3} - \frac{21}{2} y^{2} - 3 x + 30 y

f (x) = - x^{3} + 9 x^{2} + 165 x - 300

f (x) = \frac{1}{x}, 2 \leq x \leq 3

f (x) = x^{2} - 18 x + 86

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

What is the extreme f(x,y)=sqrt(400-16x^2-64y^2) ?
The extreme f(x,y)=sqrt(400-16x^2-64y^2) is Maximum(0,0)