What is the extreme f(x,y)=x^2+7y^2+x-6y+4 ?

The extreme f(x,y)=x^2+7y^2+x-6y+4 is Minimum(-1/2 , 3/7)

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+7y^2+x-6y+4

extreme points

f (x, y) = x^{2} + 7 y^{2} + x - 6 y + 4

Minimum (- \frac{1}{2}, \frac{3}{7})

Solution steps

Find the critical points:

(- \frac{1}{2}, \frac{3}{7})

\frac{\partial ^{2} f}{\partial x ^{2}} = 14

D (x, y) = 28

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, \frac{3}{7}) :

Minimum

Minimum (- \frac{1}{2}, \frac{3}{7})

f (x) = 2 x^{2} ln (x) - 19 x^{2}

f (x, y) = x^{2} + 9 x y + y^{2}

f (x) = x^{4} - 12 x^{3} + 4

f (x) = 2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11

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

What is the extreme f(x,y)=x^2+7y^2+x-6y+4 ?
The extreme f(x,y)=x^2+7y^2+x-6y+4 is Minimum(-1/2 , 3/7)