What is the extreme f(x)=(40)/r+4pir^2 ?

The extreme f(x)=(40)/r+4pir^2 is Minimum(\sqrt[3]{5/pi},8*5^{2/3}\sqrt[3]{pi}+4pi(5/pi)^{2/3})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(40)/r+4pir^2

extreme points

f (x) = \frac{40}{r} + 4 π r^{2}

Solution steps

f^{'} (r) = - \frac{40}{r ^{2}} + 8 π r

Find intervals:

Decreasing

: - \infty < r < 0,

Decreasing

Increasing

Plug

into

f (x) = x \cdot x + 2

f (x, y) = 13 - 4 x + 8 y

g (x) = x + \frac{4}{x ^{2}}

f (1, y) = y^{2} + y + 6

f (x) = - 10 x^{2} + 60 x - 80

What is the extreme f(x)=(40)/r+4pir^2 ?
The extreme f(x)=(40)/r+4pir^2 is Minimum(\sqrt[3]{5/pi},8*5^{2/3}\sqrt[3]{pi}+4pi(5/pi)^{2/3})