What is the extreme f(x,y)=x^3+8y^3-12xy ?

The extreme f(x,y)=x^3+8y^3-12xy is Saddle(0,0),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)=x^3+8y^3-12xy

extreme points

f (x, y) = x^{3} + 8 y^{3} - 12 x y

Saddle (0, 0), Minimum (2, 1)

Solution steps

Find the critical points:

(0, 0), (2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 48 y

D (x, y) = 288 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 1) :

Minimum

Saddle (0, 0), Minimum (2, 1)

f (x) = 4 x e^{- 4 x + 1}

f (x) = \frac{( x ^{2} - 9 x + 50 )}{x - 7}

f (x) = x^{4} - 32 x + 9

f (x) = - 6 x^{2} - 12 x - 9, - 3 \leq x \leq 2

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

What is the extreme f(x,y)=x^3+8y^3-12xy ?
The extreme f(x,y)=x^3+8y^3-12xy is Saddle(0,0),Minimum(2,1)