Maximum of volume

The shell of the cone is formed by winding a circular section with a radius of 1. For what central angle of a given circular section will the volume of the resulting cone be maximum?

Correct result:

A =  293.9388 °

Solution:

$$\begin{gathered} s=1 \\ V={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot h \\ h=\sqrt{s^2-r^2} \\ V={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot \sqrt{s^2-r^2} \\ V={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot \sqrt{1-r^2} \\ {V}^{\mathrm{\prime }}=0 \\ {V}^{\mathrm{\prime }}={\displaystyle \frac{dV}{dr}}={\displaystyle \frac{\pi r(2-3r^2)}{3\sqrt{1-r^2}}} \\ r>0;r\mathrm{\ne }1 \\ 2-3r^2=0 \\ r=\sqrt{2\mathrm{/}3}\doteq 0.8165 \\ {o}_1=2\pi \cdot r=2\cdot 3.1416\cdot 0.8165\doteq 5.1302 \\ {o}_2=2\pi \cdot s=2\cdot 3.1416\cdot 1\doteq 6.2832 \\ A=360\cdot {\displaystyle \frac{o_1}{o_2}}=360\cdot {\displaystyle \frac{5.1302}{6.2832}}=120\sqrt{6}=293.9388^{\circ }=293^{\circ }56^{\mathrm{\prime }}20\mathrm{"} \end{gathered}$$

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