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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 5

f (x) = x^{3} - 27 x + 5

f (x) = 6 x + \frac{384}{x}

f (x) = - 2 x^{3} + 36 x^{2} - 192 x

f (x) = 2 x^{3} - 6 x^{2} - 144 x + 7

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