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The text in the image is in Finnish, and it describes a problem related to optimizing the surface area of a tent made in the shape of a cone using a given amount of fabric. Here's the translation and interpretation of the problem:

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**Problem:**

"To create a cone-shaped tent, 16 square meters of fabric has been reserved. The fabric is not used for the base of the tent. Determine the diameter of the base of the cone when the volume of the tent is maximized. A CAS (Computer Algebra System) calculator is allowed and recommended. (1 point)"

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### Solution Approach:

This is an optimization problem involving the surface area and volume of a cone. The key details are:
- The tent is in the shape of a cone.
- The surface area (without the base) is given as 16 square meters.
- We need to maximize the volume of the cone.

For a cone:
- The lateral surface area \( A \) (without the base) is given by \( A = \pi r l \), where \( r \) is the radius of the base and \( l \) is the slant height of the cone.
- The volume of the cone \( V \) is given by \( V = \frac{1}{3} \pi r^2 h \), where \( h \) is the height of the cone.
- The slant height \( l \) can be expressed using the Pythagorean theorem as \( l = \sqrt{r^2 + h^2} \).

### Steps:
1. Express the surface area constraint: 
   \[
   \pi r l = 16
   \]
   From this equation, solve for \( l \):
   \[
   l = \frac{16}{\pi r}
   \]
2. Express the slant height in terms of \( r \) and \( h \):
   \[
   l = \sqrt{r^2 + h^2}
   \]
   Substitute the expression for \( l \) from step 1:
   \[
   \frac{16}{\pi r} = \sqrt{r^2 + h^2}
   \]
   Square both sides to eliminate the square root:
   \[
   \left( \frac{16}{\pi r} \right)^2 = r^2 + h^2
   \]
   Solve for \( h^2 \):
   \[
   h^2 = \left( \frac{16}{\pi r} \right)^2 - r^2
   \]
3. Substitute this expression for \( h^2 \) into the volume formula:
   \[
   V = \frac{1}{3} \pi r^2 h
   \]
   First, find \( h \):
   \[
   h = \sqrt{\left( \frac{16}{\pi r} \right)^2 - r^2}
   \]
   Then, substitute \( h \) into the volume equation. This gives the volume \( V \) as a function of \( r \).
4. Differentiate the volume function with respect to \( r \) and set the derivative equal to zero to find the value of \( r \) that maximizes the volume.
5. Finally, calculate the corresponding diameter \( d = 2r \).

Would you like me to proceed with the detailed calculation for this optimization problem? Let me know if you have any specific questions or need further clarifications!

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### Relative Questions:
1. What is the importance of maximizing the volume in this context?
2. How does the surface area constraint affect the shape of the cone?
3. Can you think of other applications where such optimizations are useful?
4. What would happen if the surface area included the base of the cone?
5. How can we use a CAS calculator to solve this problem more efficiently?

### Tip:
In optimization problems involving geometry, always use known formulas and relations between different dimensions to reduce the number of variables before solving.

To create a cone-shaped tent, 16 square meters of fabric has been reserved. The fabric is not used for the base of the tent. Determine the diameter of the base of the cone when the volume of the tent is maximized. A CAS (Computer Algebra System) calculator is allowed and recommended.

Learn how to determine the diameter of the base of a cone-shaped tent that maximizes its volume using 16 square meters of fabric. This problem involves key geometry and calculus concepts, including optimization techniques and the Pythagorean theorem. Suitable for students in Grades 10-12.

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