What is the extreme f(x,y)=sqrt(6-(x^2)/3-(y^2)/2) ?

The extreme f(x,y)=sqrt(6-(x^2)/3-(y^2)/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(6-(x^2)/3-(y^2)/2)

extreme points

f (x, y) = 6 - \frac{x ^{2}}{3} - \frac{y ^{2}}{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{6 ( 18 - x ^{2} )}{( 36 - 2 x ^{2} - 3 y ^{2} ) 36 - 2 x ^{2} - 3 y ^{2}}

D (x, y) = \frac{36}{( 36 - 2 x ^{2} - 3 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = x^{8} ln (x)

f (x) = 2 x^{3} + 4 x^{2}

f (x) = ln (x^{2} + 2)

- lo g_{5} (x)

8 x^{2} + 14 x y + 3 y^{2} + 10 x - 4

What is the extreme f(x,y)=sqrt(6-(x^2)/3-(y^2)/2) ?
The extreme f(x,y)=sqrt(6-(x^2)/3-(y^2)/2) is Maximum(0,0)