What is the extreme f(x,y)=sqrt(36-9x^2-4y^2) ?

The extreme f(x,y)=sqrt(36-9x^2-4y^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(36-9x^2-4y^2)

extreme points

f (x, y) = 36 - 9 x^{2} - 4 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 ( 36 - 9 x ^{2} )}{( 36 - 9 x ^{2} - 4 y ^{2} ) 36 - 9 x ^{2} - 4 y ^{2}}

D (x, y) = \frac{1296}{( 36 - 9 x ^{2} - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = (x^{2} + y^{2}) \cdot e^{- x - y}

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

f (x, y) = 4 x y - x^{4} - y^{4}

f (x, y) = x^{y}

What is the extreme f(x,y)=sqrt(36-9x^2-4y^2) ?
The extreme f(x,y)=sqrt(36-9x^2-4y^2) is Maximum(0,0)