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The problem you provided is in Vietnamese and involves calculating the minimum cost for producing rectangular tanks without a lid. Here's a translation and breakdown of the problem:

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A facility produces rectangular tanks **without a lid**. One type of tank has a volume of \( 4096 \, \text{dm}^3 \), where the length of the base is twice the width. The cost to make the base is 72 per square decimeter, and the cost for the sides is 12 per square decimeter. 

Let the width of the base be \( x \, \text{dm} \). The cost of the materials for the tank is represented by the function \( S(x) = 12 \left( ax^3 + \frac{b}{x} \right) \), where \( a \) and \( b \) need to be determined.

We need to find the correct values for \( a \), \( b \), and the minimal cost.

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### Step-by-Step Plan:

1. **Determine the dimensions of the tank**:
   - Let the width of the base be \( x \, \text{dm} \).
   - The length of the base is double the width, so it's \( 2x \, \text{dm} \).
   - The height can be represented as \( h \, \text{dm} \).

2. **Set up the volume equation**:
   The volume \( V \) of the tank is given by:
   \[
   V = \text{length} \times \text{width} \times \text{height} = 2x \times x \times h = 2x^2 h
   \]
   Since the volume is \( 4096 \, \text{dm}^3 \), we have the equation:
   \[
   2x^2 h = 4096 \quad \Rightarrow \quad h = \frac{4096}{2x^2} = \frac{2048}{x^2}
   \]

3. **Calculate the cost function**:
   - **Cost of the base**: The area of the base is \( 2x^2 \), and the cost per square decimeter for the base is 72, so the cost for the base is:
     \[
     \text{Base cost} = 72 \times 2x^2 = 144x^2
     \]
   - **Cost of the sides**: There are four sides. The areas and costs are:
     - Two sides have dimensions \( x \times h \), with an area of \( x \times h \), and a cost of \( 12 \times x \times h \).
     - Two sides have dimensions \( 2x \times h \), with an area of \( 2x \times h \), and a cost of \( 12 \times 2x \times h \).
     The total cost for the sides is:
     \[
     \text{Side cost} = 12 \times \left( 2xh + 4xh \right) = 72xh = 72 \times \frac{2048}{x}
     \]

4. **Total cost function**:
   The total cost function is the sum of the base and side costs:
   \[
   S(x) = 144x^2 + 72 \times \frac{2048}{x} = 144x^2 + \frac{147456}{x}
   \]

This matches the structure \( S(x) = 12 \left( ax^3 + \frac{b}{x} \right) \), where:
\[
a = 12 \quad \text{and} \quad b = 12288
\]

5. **Minimizing the cost**:
   To find the value of \( x \) that minimizes the cost, take the derivative of \( S(x) \) with respect to \( x \) and set it to zero:
   \[
   S'(x) = 288x - \frac{147456}{x^2}
   \]
   Setting this equal to zero:
   \[
   288x = \frac{147456}{x^2} \quad \Rightarrow \quad x^3 = \frac{147456}{288} = 512 \quad \Rightarrow \quad x = 8 \, \text{dm}
   \]

6. **Calculating the minimum cost**:
   Substitute \( x = 8 \, \text{dm} \) into the cost function \( S(x) \):
   \[
   S(8) = 144 \times 8^2 + \frac{147456}{8} = 9216 + 18432 = 27648
   \]
   The minimum cost is 27,648 units.

---

### Final Answer:

- The correct values are \( a = 12 \), \( b = 12288 \), and the minimal cost is **27,648** units.

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Let me know if you'd like further clarifications or details!

Here are 5 related questions for practice:
1. How would the problem change if the base of the tank had a square shape?
2. What would happen if the height of the tank was fixed instead of the volume?
3. How does changing the cost per square decimeter affect the overall cost function?
4. What is the impact on the cost function if the dimensions of the base are not proportional?
5. Can the optimization be performed using different methods, such as Lagrange multipliers?

**Tip**: When optimizing costs or any other function, always ensure that the constraints are included in your calculations to avoid errors.

Một cơ sở chuyên sản xuất các bể chứa dạng hình hộp chữ nhật không tính nắp. Một loại bể có thể tích 4096 dm³, đáy có chiều dài gấp đôi chiều rộng. Chi phí làm mặt đáy có giá 72 và làm các mặt bên có giá 12; đơn vị tính là nghìn đồng / 1 dm². Đặt chiều rộng đáy bể là x dm, chi phí nguyên liệu làm bể được tính bởi hàm S(x) có dạng S(x) = 12(ax³ + b/x). Tìm a, b, t và chi phí thấp nhất để làm bể theo yêu cầu.

Learn how to minimize the cost of a rectangular tank with a volume of 4096 dm³, where the base length is twice the width. This problem involves calculating the minimum cost using the cost function S(x) = 12(ax³ + b/x) and includes step-by-step solutions for finding the values of a, b, and the minimal cost.

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