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

Minimum (3 \frac{1.19685 E 15}{2.0 E 14}, 1491.13746 \dots)

Solution steps

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

Find intervals:

Decreasing

: - \infty < r < 0,

Decreasing

: 0 < r < 3 \frac{1.19685 E 15}{2.0 E 14},

Increasing

: 3 \frac{1.19685 E 15}{2.0 E 14} < r < \infty

Plug

r = 3 \frac{1.19685 E 15}{2.0 E 14}

into

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

Minimum (3 \frac{1.19685 E 15}{2.0 E 14}, 1491.13746 \dots)

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…)