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This problem deals with optimizing the cost of constructing a cylindrical can, which has a capacity of one-quarter liter (0.25 liters). Let's break it down step by step to find the dimensions of the can with the lowest cost.

### Key Details:
1. **Volume of the Cylinder**:
   - The volume \( V \) of the cylinder is given by:
     \[
     V = \pi r^2 h
     \]
   - Since the volume must be 0.25 liters (or 250 cm³), we have the constraint:
     \[
     \pi r^2 h = 250 \, \text{cm}^3
     \]

2. **Cost Information**:
   - The cost for the material used in the top and bottom (circular parts) is 3 pesos per square centimeter.
   - The cost for the material used for the lateral surface is 2 pesos per square centimeter.

3. **Surface Area**:
   - The surface area \( A \) of the can consists of:
     - The area of the two circular ends: \( 2 \pi r^2 \)
     - The area of the lateral surface: \( 2 \pi r h \)
   - Thus, the total surface area is:
     \[
     A = 2 \pi r^2 + 2 \pi r h
     \]

4. **Cost Function**:
   - The total cost \( C \) is based on the surface area and the cost per square centimeter:
     \[
     C = \text{Cost of tops and bottoms} + \text{Cost of lateral area}
     \]
     \[
     C = 3(2 \pi r^2) + 2(2 \pi r h) = 6 \pi r^2 + 4 \pi r h
     \]
   - From the volume constraint \( \pi r^2 h = 250 \), we can solve for \( h \):
     \[
     h = \frac{250}{\pi r^2}
     \]
   - Substitute this into the cost function to get the cost as a function of \( r \):
     \[
     C(r) = 6 \pi r^2 + 4 \pi r \left( \frac{250}{\pi r^2} \right)
     \]
     Simplifying:
     \[
     C(r) = 6 \pi r^2 + \frac{1000}{r}
     \]

### Optimization:
Now, we need to minimize the cost function \( C(r) \). To do this, we take the derivative of \( C(r) \) with respect to \( r \) and set it equal to zero:

\[
\frac{dC}{dr} = 12 \pi r - \frac{1000}{r^2}
\]
Setting the derivative equal to zero:
\[
12 \pi r = \frac{1000}{r^2}
\]
Multiplying both sides by \( r^2 \) and solving for \( r \):
\[
12 \pi r^3 = 1000
\]
\[
r^3 = \frac{1000}{12 \pi}
\]
\[
r = \sqrt[3]{\frac{250}{3 \pi}} \approx 3.842 \, \text{cm}
\]

### Finding the Height \( h \):
Using the volume equation \( \pi r^2 h = 250 \), we can find \( h \):
\[
h = \frac{250}{\pi (3.842)^2} \approx 5.387 \, \text{cm}
\]

### Summary:
- **Radius (r)**: \( 3.842 \, \text{cm} \)
- **Height (h)**: \( 5.387 \, \text{cm} \)

These are the dimensions of the can that minimize the cost.

---

Would you like further details on the optimization process, or should I help with constructing the graph in GeoGebra?

### Related Questions:
1. How does the cost of the cylinder change if the price per square centimeter of the lateral surface changes?
2. How do you find the maximum or minimum values for functions like the cost function?
3. How would the problem change if the volume constraint were different?
4. What happens to the dimensions if the material cost of the tops is reduced?
5. How can this problem be modeled using calculus in more complex shapes?

### Tip:
Always check if constraints like volume are met when solving optimization problems. It ensures that the solution is physically valid.

Una lata cilíndrica debe tener una capacidad de un cuarto de litro. El costo del material en las tapas es de tres pesos por centímetro cuadrado y el de la parte lateral es de dos pesos por centímetro cuadrado. ¿Cuáles son las dimensiones de la lata de menor costo?

This problem involves optimizing the construction cost of a cylindrical can with a capacity of 250 cm³. Learn how to minimize costs using volume and surface area formulas, along with optimization techniques in calculus. Suitable for students in Grades 10-12.

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