What is the extreme f(x,y)=x^2+xy+y^2+y ?

The extreme f(x,y)=x^2+xy+y^2+y is Minimum(1/3 ,-2/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=x^2+xy+y^2+y

extreme points

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

Minimum (\frac{1}{3}, - \frac{2}{3})

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{2}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{2}{3}) :

Minimum

Minimum (\frac{1}{3}, - \frac{2}{3})

f (x) = x^{4} - 4 x^{3} - 18 x^{2}

y = x^{4} - 2 x^{2}

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

f (x, y) = ln (x^{2} + y^{2} - 1)

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

What is the extreme f(x,y)=x^2+xy+y^2+y ?
The extreme f(x,y)=x^2+xy+y^2+y is Minimum(1/3 ,-2/3)