1. A company produces cylindrical tiles. Each tile has a diameter of $6 \mathrm{~cm}$ and a volume of $600 \mathrm{~cm}^{3}$. The company decides to change the diameter of the tiles so that they will have a volume of $700 \mathrm{~cm}^{3}$ and the same height as before. How many $\mathrm{cm}^{2}$ of tile material does the company use to produce the modified tile?

(6 points)
1. Two cans have the same height. From the formula for the volume of a cylinder $V=\pi r^{2} \cdot v$ we express the height $v=\frac{V}{\pi \cdot r^{2}}$. We equate the heights of both cylinders and from the relation $\frac{V_{1}}{\pi r_{1}^{2}}=\frac{V_{2}}{\pi r_{2}^{2}}$ we calculate the radius of the second can $r_{2} \doteq 3.24 \mathrm{~cm}$. We calculate the surface area of the modified can, which is $P=2 \pi r_{2}^{2}+2 \pi r_{2} \cdot v \doteq 2 \pi \cdot 3.24+2 \pi \cdot 3.24 \cdot 21.22 \doteq 498$.

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-11.jpg?height=69&width=1642&top_left_y=545&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-11.jpg?height=55&width=1639&top_left_y=595&top_left_x=206)

Considering or noting that the heights of both cans are equal .........................................

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-11.jpg?height=100&width=1642&top_left_y=692&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-11.jpg?height=55&width=1642&top_left_y=775&top_left_x=204)

Calculating the surface area of the modified can $P=2 \pi r_{2}^{2}+2 \pi r_{2} \cdot v \doteq 498 \mathrm{~cm}^{2} \ldots \ldots .1$ point NOTE: If the units are missing in the result, we deduct 1 point.