What is the extreme f(x,y)=12xy-x^2-3y^2 ?

The extreme f(x,y)=12xy-x^2-3y^2 is Saddle(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=12xy-x^2-3y^2

extreme points

f (x, y) = 12 x y - x^{2} - 3 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = - 132

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{3} - 12 x^{2} + 36 x - 3, 0 \leq x \leq 9

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

f (x) = 2 x^{3} - 21 x^{2} + 72 x - 8

\frac{1}{x ^{2} + x + 1}

f (x) = e^{14 x} + e^{- x}

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