What is the extreme f(x)=2pir^2+(500)/r ?

The extreme f(x)=2pir^2+(500)/r is Minimum(5/(\sqrt[3]{pi)},50pi^{1/3}+100\sqrt[3]{pi})

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extreme f(x)=2pir^2+(500)/r

extreme points

f (x) = 2 π r^{2} + \frac{500}{r}

Solution steps

f^{'} (r) = 4 π r - \frac{500}{r ^{2}}

Find intervals:

Decreasing

: - \infty < r < 0,

Decreasing

Increasing

Plug

into

\frac{y ( y - 5 )}{y ^{2} - y + 1}

f (x) = \frac{x + 4}{1 - x}

f (x) = - e^{(- x)} + e^{x}

\frac{- 3 x}{2 x + 5}

\frac{1}{2} x - 30

What is the extreme f(x)=2pir^2+(500)/r ?
The extreme f(x)=2pir^2+(500)/r is Minimum(5/(\sqrt[3]{pi)},50pi^{1/3}+100\sqrt[3]{pi})