## Task 33/64

For the approximate calculation of the volume of bodies in the form of truncated pyramids and cones, which often occur in technology, the rule of thumb

$$
V \approx V_{N}=\frac{h}{2}\left(G_{1}+G_{2}\right)
$$

is often applied. Here, $h$ denotes the height of the truncated body, and $G_{1}$ and $G_{2}$ denote the areas of the two parallel boundary surfaces that are perpendicular to the height $h$.

When there are large percentage differences between the parallel boundary surfaces, the error in using this formula can be significant (up to 50%). Therefore, solving the following problem is important:

By how many percent may the difference in the sizes of $G_{1}$ and $G_{2}$ at most be, so that the error of the approximation formula does not exceed 1% or $n\%$?
The error in using the rule of thumb is

$$
f=V_{N}-V=\frac{h}{2}\left(G_{1}+G_{2}\right)-\frac{h}{3}\left(G_{1}+G_{2}+\sqrt{G_{1} G_{2}}\right)=\frac{h}{6}\left(G_{1}+G_{2}-2 \sqrt{G_{1} G_{2}}\right)=\frac{h}{6}\left(\sqrt{G_{1}}-\sqrt{G_{2}}\right)^{2}
$$

From this, it follows that the rule of thumb always yields a value that is too large for $G_{1} \neq G_{2}$. The further transformation

$$
f=\frac{h}{6} G_{1}\left(1-\sqrt{\frac{G_{2}}{G_{1}}}\right)^{2}
$$

shows that the error $f$ is proportional to the height $h$ and to $G_{1}$ or $G_{2}$ when the ratio $G_{2}: G_{1}$ is constant.

For the relative error $\frac{f}{V}$, one obtains

$$
\frac{f}{V}=\frac{\left(\sqrt{G_{1}}-\sqrt{G_{2}}\right)^{2}}{2\left(G_{1}+G_{2}+\sqrt{G_{1} G_{2}}\right.}=\left(1-\frac{G_{2}}{G_{1}}\right)^{2}: 2\left(1+\frac{G_{2}}{G_{1}}+\sqrt{\frac{G_{2}}{G_{1}}}\right)
$$

If $\sqrt{\frac{G_{2}}{G_{1}}}$ is replaced by $x$ (where $x$ is the ratio of corresponding lengths in the parallel boundary surfaces), then

$$
\frac{f}{V}=\frac{(1-x)^{2}}{2\left(1+x+x^{2}\right)}
$$

The relative error thus depends only on the ratio $G_{2}: G_{1}$. If it is to be less than or at most $0.01=1 \%$, the inequality for $x$ is

$$
(1-x)^{2} \leq 0.01 \cdot 2\left(1+x+x^{2}\right)
$$

since $2\left(1+x+x^{2}\right)>0$. Through appropriate transformations, it follows that

$$
50(1-x)^{2} \leq 1+x+x^{2} \rightarrow x^{2}-\frac{101}{49} x \leq-1 \rightarrow\left(x-\frac{101}{98}\right)^{2}<\frac{597}{98^{2}}
$$

Let now (without loss of generality) $G_{2} \leq G_{1}$, so $x \leq 1$. Then one obtains (with slide rule accuracy)

$$
x \geq \frac{101-\sqrt{597}}{98}=0.781 \rightarrow 0.610 \leq x \leq 1
$$

Thus, $G_{2} \geq 0.610 G_{1}$, meaning $G_{2}$ must be at least $61.0 \%$ of $G_{1}$ if the error in using the rule of thumb is not to exceed $1 \%$. The difference $\left|G_{1}-G_{2}\right|$ must not exceed $39 \%$ of the larger area.

If the error is not to exceed $n \%$, the corresponding inequality is

$$
(1-x)^{2} \leq \frac{2 n}{100}\left(1+x+x^{2}\right)
$$

This results in the calculation performed for $n=1$, where $\frac{100+n}{50-n}=k$ is set and it is noted that $k \geq 2$ or $\frac{k}{2} \geq 1$:

$$
x \geq \frac{k+\sqrt{k^{2}-4}}{2}
$$

An investigation of the error function

$$
y=\frac{f}{V}=\frac{(1-x)^{2}}{2\left(1+x+x^{2}\right)}
$$

yields, in the interval $0 \leq x \leq 1$, the maximum value $0.5=50 \%$, and this value is assumed when the body is a pyramid or a cone $\left(G_{2}=0\right)$.