Every Sunday, Pedro prepares cannelloni for lunch. First, he cuts rectangles of pasta measuring 16 by $12 \mathrm{~cm}$, and then he glues the two longer sides, overlapping a strip of $2 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-033.jpg?height=237&width=417&top_left_y=915&top_left_x=1462)

This way, he obtains cylinders that he fills with ricotta, using $500 \mathrm{~g}$ of ricotta per cylinder. On a beautiful Sunday, with the same number of pasta rectangles measuring 16 by $12 \mathrm{~cm}$, he decides to produce the cylinders by gluing the shorter sides, always overlapping a strip of $2 \mathrm{~cm}$. In this situation, will he use more or less ricotta than before? How much?
By gluing the rectangles of dough along the longer side, Pedro obtains a cylinder with a circular base that is $10 \mathrm{~cm}$ in length and $16 \mathrm{~cm}$ in height. The volume he then fills with ricotta is given by $V=$ area of the base $\times$ height of the cylinder. The area of the base is given by $\pi \times r^{2}$, where $r$ denotes the radius of the base. Let's calculate the radius, knowing that the perimeter of the base is $10 \mathrm{~cm}$. We have $2 \pi r=10$, or $r=5 / \pi$. Thus, the volume of ricotta for each cannelloni is given, in this case, by

$$
V=\pi \times \frac{5^{2}}{\pi^{2}} \times 16=\frac{16 \times 25}{\pi}=\frac{400}{\pi} \mathrm{cm}^{3}
$$

Now, by gluing the rectangles of dough along the shorter side, Pedro obtains a cylinder with a circular base that is $14 \mathrm{~cm}$ in perimeter and $12 \mathrm{~cm}$ in height.

The radius of the base is now given by $r^{\prime}=14 / 2 \pi=7 / \pi$. Thus, the volume of ricotta for each cannelloni is given, in this case, by

$$
V^{\prime}=\pi \times \frac{7^{2}}{\pi^{2}} \times 12=\frac{588}{\pi} \mathrm{cm}^{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-113.jpg?height=317&width=548&top_left_y=658&top_left_x=1325)

Finally, to calculate the new cost of ricotta, we use a direct proportion.

$$
\begin{array}{ccc}
\text { Volume }\left(\mathrm{cm}^{3}\right) & & \text { Ricotta }(\mathrm{g}) \\
\frac{400}{\pi} & \longrightarrow & 500 \\
\frac{588}{\pi} & \longrightarrow & x
\end{array}
$$

It follows that

$$
x=\frac{500 \times 588}{400}=735 \mathrm{~g},
$$

so now Pedro spends $235 \mathrm{~g}$ more ricotta per cannelloni.