Three cylinders have heights and base radii equal to $10 \mathrm{~cm} \times 10 \mathrm{~cm}, 10 \mathrm{~cm} \times 5 \mathrm{~cm}$, and $20 \mathrm{~cm} \times 5 \mathrm{~cm}$, and volumes $V_{1}, V_{2}$, and $V_{3}$, respectively.
![](https://cdn.mathpix.com/cropped/2024_05_01_df4451ddcede6035446dg-2.jpg?height=538&width=644&top_left_y=390&top_left_x=817)

(a) Write in ascending order the volumes $V_{1}, V_{2}$, and $V_{3}$ of the three cylinders.

(b) Give the dimensions of a cylinder $V_{4}$ whose volume is between $V_{2}$ and $V_{3}$.

(c) Give the dimensions of a cylinder $V_{5}$ whose volume is between $V_{1}$ and $V_{3}$.
(a) Given that the volume of a cylinder with radius $R$ and height $h$ is $\pi R^{2} h$, we have that the volumes $V_{1}, V_{2}$, and $V_{3}$ are:

$V_{1}=\pi \times 10^{3}=1000 \pi, V_{2}=\pi \times 5^{2} \times 10=250 \pi, \quad V_{3}=\pi \times 5^{2} \times 20=500 \pi$

Thus, we have $V_{1}>V_{3}>V_{2}$.

(b) Since the two cylinders have the same radius, it is sufficient to keep the radius of the cylinder at $5 \mathrm{~cm}$ and the height between $10 \mathrm{~cm}$ and $20 \mathrm{~cm}$, for example: $h=15 \mathrm{~cm}$. In this case, the volume $V_{4}$ is: $\pi \times 5^{2} \times 15=375 \pi \mathrm{cm}^{3}$.

(c) To construct a cylinder with volume $V_{5}$ between $V_{1}$ and $V_{3}$, we can reduce the radius of the cylinder with volume $V_{5}$ to $8 \mathrm{~cm}$ and take the height as $10 \mathrm{~cm}$, the smaller of the two heights, obtaining a cylinder with volume $\pi \times 8^{2} \times 10=640 \pi \mathrm{cm}^{3}$.