What is the extreme f(x)=48pir^2+(1804.8)/r ?

The extreme f(x)=48pir^2+(1804.8)/r is Minimum(\sqrt[3]{(1.19685E15)/(2.0E14)},1491.13746…)

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extreme f(x)=48pir^2+(1804.8)/r

extreme points

f (x) = 48 π r^{2} + \frac{1804.8}{r}

Solution steps

f^{'} (r) = 301.59289 \dots r - \frac{1804.8}{r ^{2}}

Find intervals:

Decreasing

: - \infty < r < 0,

Decreasing

Increasing

Plug

into

48 π r^{2} + \frac{1804.8}{r} : 1491.13746 \dots

f (x) = \frac{x + 1}{x ^{2} - 4}

f (x) = (x - 4) ln (x - 4)

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

f (x, y) = 5 x^{2} + x y^{2} + y^{2}

f (x) = 1 + 12 x - x^{3}

What is the extreme f(x)=48pir^2+(1804.8)/r ?
The extreme f(x)=48pir^2+(1804.8)/r is Minimum(\sqrt[3]{(1.19685E15)/(2.0E14)},1491.13746…)