What is the extreme f(x,y)=ln(9x^2+3y^2-4) ?

The extreme f(x,y)=ln(9x^2+3y^2-4) 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)=ln(9x^2+3y^2-4)

extreme points

f (x, y) = ln (9 x^{2} + 3 y^{2} - 4)

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{6 ( - 3 y ^{2} + 9 x ^{2} - 4 )}{( 9 x ^{2} + 3 y ^{2} - 4 ) ^{2}}

D (x, y) = \frac{- 8748 x ^{4} - 5832 x ^{2} y ^{2} - 972 y ^{4} + 1728}{( - 4 + 9 x ^{2} + 3 y ^{2} ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 3 x^{\frac{2}{3}} - 7

f (x, y) = \frac{1}{x + y + 3}

f (x, y) = x y + x^{2} y + x y^{2}

\frac{x ^{2} + 1}{x ^{3} - 1}

f (x) = - 17 \frac{x}{x ^{2} + y ^{2} + 1}

What is the extreme f(x,y)=ln(9x^2+3y^2-4) ?
The extreme f(x,y)=ln(9x^2+3y^2-4) is Maximum(0,0)