What is the extreme P(x,y)=x+y^2-10x^2y^2+9y^6 ?

The extreme P(x,y)=x+y^2-10x^2y^2+9y^6 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme P(x,y)=x+y^2-10x^2y^2+9y^6

extreme

Saddle \frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, \frac{- 10 + 37}{4 3 ^{3} 8 40}, Saddle \frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, - \frac{- 10 + 37}{4 3 ^{3} 8 40}

Solution steps

Find the critical points:

\frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, \frac{- 10 + 37}{4 3 ^{3} 8 40}, \frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, - \frac{- 10 + 37}{4 3 ^{3} 8 40}

\frac{\partial ^{2} f}{\partial x ^{2}} = 270 y^{4} + 2 - 20 x^{2}

D (x, y) = - 5400 y^{6} - 40 y^{2} - 1200 x^{2} y^{2}

Check the sign of at each critical point

Check critical point

\frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, \frac{- 10 + 37}{4 3 ^{3} 8 40} :

Saddle

Check critical point

\frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, - \frac{- 10 + 37}{4 3 ^{3} 8 40} :

Saddle

Saddle \frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, \frac{- 10 + 37}{4 3 ^{3} 8 40}, Saddle \frac{3 3}{2 \cdot 5 ^{\frac{3}{4}} 4 2 37 - 10}, - \frac{- 10 + 37}{4 3 ^{3} 8 40}

What is the extreme P(x,y)=x+y^2-10x^2y^2+9y^6 ?
The extreme P(x,y)=x+y^2-10x^2y^2+9y^6 is