What is the extreme f(x,y)=sqrt(400-49x^2-9y^2) ?

The extreme f(x,y)=sqrt(400-49x^2-9y^2) is Maximum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=sqrt(400-49x^2-9y^2)

extreme points

f (x, y) = 400 - 49 x^{2} - 9 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{9 ( 400 - 49 x ^{2} )}{( 400 - 49 x ^{2} - 9 y ^{2} ) 400 - 49 x ^{2} - 9 y ^{2}}

D (x, y) = \frac{176400}{( 400 - 49 x ^{2} - 9 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x^{3} + 24 x^{2} + 72 x + 7

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

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

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

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

What is the extreme f(x,y)=sqrt(400-49x^2-9y^2) ?
The extreme f(x,y)=sqrt(400-49x^2-9y^2) is Maximum(0,0)