What is the extreme f(x)=9-2x+4y-x^2-4y^2 ?

The extreme f(x)=9-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=9-2x+4y-x^2-4y^2

extreme points

f (x) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

Maximum (- 1, \frac{1}{2})

Solution steps

Find the critical points:

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

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

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

Maximum

Maximum (- 1, \frac{1}{2})

f (x) = x^{3} + 2 x^{2} + 1

\frac{5 x ^{2}}{x ^{2} + 16}

f (x) = 8 x - x^{4}, - 2 \leq x \leq 1

f (x, y) = 2 x^{2} + 4 x y^{2} - 2 x y - 10 x - 2 y + 2

f (x) = 3 (x - 4) (x + 7)^{2}

What is the extreme f(x)=9-2x+4y-x^2-4y^2 ?
The extreme f(x)=9-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)