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

The extreme f(x,y)=x-4x^2y+y^2 is Saddle(1/(4^{2/3)}, 1/(2\sqrt[3]{4)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=x-4x^2y+y^2

extreme

Saddle (\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

Solution steps

Find the critical points:

(\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

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

D (x, y) = - 16 y - 64 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4}) :

Saddle

Saddle (\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

What is the extreme f(x,y)=x-4x^2y+y^2 ?
The extreme f(x,y)=x-4x^2y+y^2 is Saddle(1/(4^{2/3)}, 1/(2\sqrt[3]{4)})