What is the extreme f(x,y)=5x^2+5y^2+20x-10y+40 ?

The extreme f(x,y)=5x^2+5y^2+20x-10y+40 is Minimum(-2,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=5x^2+5y^2+20x-10y+40

extreme points

f (x, y) = 5 x^{2} + 5 y^{2} + 20 x - 10 y + 40

Minimum (- 2, 1)

Solution steps

Find the critical points:

(- 2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 100

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Minimum

Minimum (- 2, 1)

f (x) = x (27 - 41 + 2 x) (\frac{41}{2} - x)

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x + 2

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x + 1

f (x) = (x + 1)^{7} - 7 x - 3

f (x) = - x^{2} + 4 x y - 5 y^{2} - 6 x + 22 y + 9

What is the extreme f(x,y)=5x^2+5y^2+20x-10y+40 ?
The extreme f(x,y)=5x^2+5y^2+20x-10y+40 is Minimum(-2,1)