What is the extreme f(x,y)=x^{(2)}-2xy+4y^{(2)}-4x-2y+24 ?

The extreme f(x,y)=x^{(2)}-2xy+4y^{(2)}-4x-2y+24 is Minimum(3,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=x^{(2)}-2xy+4y^{(2)}-4x-2y+24

extreme points

f (x, y) = x^{(2)} - 2 x y + 4 y^{(2)} - 4 x - 2 y + 24

Minimum (3, 1)

Solution steps

Find the critical points:

(3, 1)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(3, 1) :

Minimum

Minimum (3, 1)

f (x) = \frac{2 + x}{7 - x}, - 7 \leq x \leq 0

y = 3 x^{4} + 16 x^{3} = x^{3} (3 x + 16)

f (x) = - 4 x^{3} - 6 x^{2} + 72 x + 6

y = 12 x^{\frac{5}{7}} + 5 x^{\frac{12}{7}}

y = x^{3} - 3 x^{2} + 7 x - 10

What is the extreme f(x,y)=x^{(2)}-2xy+4y^{(2)}-4x-2y+24 ?
The extreme f(x,y)=x^{(2)}-2xy+4y^{(2)}-4x-2y+24 is Minimum(3,1)