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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=-x^2-y^2+8x+8y

extreme points

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

Maximum (4, 4)

Solution steps

Find the critical points:

(4, 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(4, 4) :

Maximum

Maximum (4, 4)

f (x) = sin (3 x), - \frac{π}{4} \leq x \leq \frac{π}{3}

f (x, y) = x^{3} + 8 x^{2} y^{2} - 8 y^{3} - x + y

f (t) = 4 t^{3} - 36 t^{2} + 10, t \geq 0

f (x) = 3 x^{4} + 20 x^{3} - 36 x^{2} - 4

2 x^{3} - 30 x^{2}, - 1 \leq x \leq 11

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