\section*{Task 20 - V01220}

Calculate the internal dimensions of a cylindrical cast iron pan with a capacity of 20 t, which should exhibit the lowest possible heat losses through appropriate shaping!

Based on experience, it is assumed that the heat losses from the surface of the liquid cast iron (per unit area) are twice the heat losses through the wall and bottom surface. (Weight of liquid cast iron: \(7.2 \mathrm{t} / \mathrm{m}^{3}\) )
}

With a density of \(7.2 \frac{t}{m^{3}}\) for the raw iron, this material occupies a cylinder with a volume of \(V=\frac{20}{7.2} m^{3}\). Let \(r>0\) be the inner radius of the raw iron pan measured in meters, and \(h>0\) be the height of the liquid raw iron, then \(\pi \cdot r^{2} \cdot h=\frac{20}{7.2}\) or equivalently \(h=\frac{20}{7.2 \cdot \pi \cdot r^{2}}\). The heat loss, which is proportional to the surface area of the cylinder formed by the raw iron, is to be minimized, with the top surface area being counted twice. Therefore, the term \(\pi \cdot (2 r h + 3 r^{2})\) or equivalently the term \(3 r^{2} + 2 r h\) needs to be minimized. Substituting the previously obtained condition for \(r\), the term to be minimized is thus

\[
f(r)=3 \cdot r^{2} + 2 \cdot \frac{20}{7.2 \cdot \pi} \cdot r^{-1}
\]

This function of \(r\) has the derivative

\[
f^{\prime}(r)=6 r - 2 \cdot \frac{20}{7.2 \cdot \pi} \cdot r^{-2}
\]

which vanishes exactly for those \(r>0\) for which the equation

\[
3 r^{3}=\frac{20}{7.2 \cdot \pi} \quad \text{or} \quad r=\left(\frac{20}{3 \cdot 7.2 \cdot \pi}\right)^{\frac{1}{3}} \approx 0.6655
\]

holds. A boundary consideration with \(r \rightarrow 0\) or \(r \rightarrow \infty\) shows that \(f(r)\) tends to infinity in both cases, so it must have a global minimum at the only critical point.

Thus, the raw iron pan has an inner radius of \(r \approx 0.6655 \, \text{m}\) and an inner height of \(h=\frac{20}{7.2 \cdot \pi \cdot r^{2}} \approx\) \(6.272 \, \text{m}\).