What is the extreme P(x,y)=x^2+5xy+4y^2+2x+5y+1 ?

The extreme P(x,y)=x^2+5xy+4y^2+2x+5y+1 is Saddle(-1,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme P(x,y)=x^2+5xy+4y^2+2x+5y+1

extreme points

P (x, y) = x^{2} + 5 x y + 4 y^{2} + 2 x + 5 y + 1

Saddle (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Saddle (- 1, 0)

f (x) = 2 x^{3} - 24 x, - 4 \leq x \leq 4

\frac{1}{4} x^{4} + \frac{1}{2} x^{3} - 3 x^{2} + 8

\frac{1}{4} x^{4} + \frac{1}{2} x^{3} - 3 x^{2} + 2

f (x) = x (25 - 39 + 2 x) (\frac{39}{2} - x)

f (x) = x^{2} + 8 x + 11

What is the extreme P(x,y)=x^2+5xy+4y^2+2x+5y+1 ?
The extreme P(x,y)=x^2+5xy+4y^2+2x+5y+1 is Saddle(-1,0)