Bottles of mineral water with a capacity of one and a half liters are tapered in an oval shape to make them easier to hold. The normal circumference of the bottle is $27.5 \mathrm{~cm}$, while at the waist - which is a $1 \mathrm{~cm}$ high cylindrical section - it is only $21.6 \mathrm{~cm}$. The cylindrical sections with different circumferences are connected above and below the waist by conical frustum sections, each $2 \mathrm{~cm}$ high. How much taller are such bottles compared to their counterparts of the same volume but with a normal circumference and no waist?
Solution. Let the radius of the larger circle of the frustum be $R$, and the radius of the smaller circle be $r$. It is sufficient to examine how much the height changes if we replace the ring-shaped part with a cylinder of equal volume and radius $R$. Let the height of this cylinder be $m$.
![](https://cdn.mathpix.com/cropped/2024_05_02_c999bb89885f04bdf3dbg-1.jpg?height=224&width=610&top_left_y=310&top_left_x=752)

The volume of the ring-shaped part:

$$
V_{1}=2 \cdot \frac{2 \pi}{3} \cdot\left(R^{2}+R r+r^{2}\right)+r^{2} \pi
$$

The volume of the cylinder: $V_{2}=R^{2} \pi m$, and since $V_{1}=V_{2}$:

$$
\frac{4 \pi}{3} \cdot\left(R^{2}+R r+r^{2}\right)+r^{2} \pi=R^{2} \pi m
$$

Simplify by $\pi$ and solve for $m$:

$$
m=\frac{\frac{4}{3} R^{2}+\frac{4}{3} R r+\frac{4}{3} r^{2}+r^{2}}{R^{2}}=\frac{4}{3}+\frac{7}{3} \cdot \frac{r^{2}}{R^{2}}+\frac{4}{3} \cdot \frac{r}{R}
$$

We know that $2 R \pi=27.5$ and $2 r \pi=21.6$, from which $\frac{r}{R}=\frac{21.6}{27.5}$ follows; substitute this into (1):

$$
m=\frac{4}{3}+\frac{7}{3} \cdot\left(\frac{21.6}{27.5}\right)^{2}+\frac{4}{3} \cdot \frac{21.6}{27.5} \approx 3.8201
$$

The height difference $d \approx 5-3.8201 \approx 1.18 \text{ cm}$: this is how much higher the "narrowed" bottles are.