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}

Minimum (\frac{5}{3 π}, 50 π^{\frac{1}{3}} + 100 3 π)

Solution steps

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

Find intervals:

Decreasing

: - \infty < r < 0,

Decreasing

: 0 < r < \frac{5}{3 π},

Increasing

: \frac{5}{3 π} < r < \infty

Plug

r = \frac{5}{3 π}

into

2 π r^{2} + \frac{500}{r} : 50 π^{\frac{1}{3}} + 100 3 π

Minimum (\frac{5}{3 π}, 50 π^{\frac{1}{3}} + 100 3 π)

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

(33, 75), (3, - 5)

\frac{x}{x ^{2} - 5 x - 14}

x - 2 y = 3

f (x) = 4 - 7 x

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})