What is the extreme f(x,y)=8x^2+8xy+10y^2-24x-28y+27 ?

The extreme f(x,y)=8x^2+8xy+10y^2-24x-28y+27 is Minimum(1,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=8x^2+8xy+10y^2-24x-28y+27

extreme points

f (x, y) = 8 x^{2} + 8 x y + 10 y^{2} - 24 x - 28 y + 27

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 20

D (x, y) = 256

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x, y) = 3 x + 2 y

f (x, y) = 3 x + 4 y

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

f (x, y) = x^{2} + x y + y^{2} - 3 x + 2

x^{4} - 4 x^{3} + 4 x^{2}

What is the extreme f(x,y)=8x^2+8xy+10y^2-24x-28y+27 ?
The extreme f(x,y)=8x^2+8xy+10y^2-24x-28y+27 is Minimum(1,1)