What is the extreme f(x,y)=x^2+4y^2-2x+8y-1 ?

The extreme f(x,y)=x^2+4y^2-2x+8y-1 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)=x^2+4y^2-2x+8y-1

extreme points

f (x, y) = x^{2} + 4 y^{2} - 2 x + 8 y - 1

Minimum (1, - 1)

Solution steps

Find the critical points:

(1, - 1)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Minimum (1, - 1)

x^{3} + 6 x^{2} - 15 x

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

f (x) = x^{3} + 15 x^{2}

f (x, y) = 3 x^{4} - 2 x y^{2} + 4 y^{3} + 1

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

What is the extreme f(x,y)=x^2+4y^2-2x+8y-1 ?
The extreme f(x,y)=x^2+4y^2-2x+8y-1 is Minimum(1,-1)