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To solve this problem, let's go step-by-step.

1. **Given information:**
   - The box has a square base with side length \( x \).
   - The height of the box is \( y \).
   - The box has an open top.
   - The volume of the box is given as \( 78732 \, \text{cm}^3 \).

2. **Objective:**
   - Find the dimensions of the box that minimize the surface area (material used) of the box.
   - First, we need to express the surface area \( A(x) \) in terms of \( x \) only.

### Step 1: Express the Volume Constraint
Since the box has a square base and height \( y \), the volume \( V \) is:
\[
V = x^2 \cdot y
\]
Given that \( V = 78732 \, \text{cm}^3 \), we can write:
\[
x^2 \cdot y = 78732
\]
Solving for \( y \) in terms of \( x \):
\[
y = \frac{78732}{x^2}
\]

### Step 2: Set Up the Surface Area Function
The surface area \( A \) of the box, with an open top, includes:
- The area of the square base: \( x^2 \).
- The area of the four sides: \( 4 \cdot x \cdot y \).

So, the surface area \( A(x) \) in terms of \( x \) and \( y \) is:
\[
A(x) = x^2 + 4xy
\]
Substitute \( y = \frac{78732}{x^2} \) into this equation:
\[
A(x) = x^2 + 4x \cdot \frac{78732}{x^2}
\]
Simplify:
\[
A(x) = x^2 + \frac{314928}{x}
\]

### Final Answer for \( A(x) \):
\[
A(x) = x^2 + \frac{314928}{x}
\]

This is the formula for the surface area in terms of \( x \). To minimize the material used, we would take the derivative of \( A(x) \) with respect to \( x \) and set it to zero to find the critical points, then check which value of \( x \) minimizes \( A(x) \).

Would you like to continue with the optimization to find the exact dimensions that minimize the surface area? 

Here are some additional questions to consider:

1. How do you find the minimum value of a function?
2. Why is it important to express \( y \) in terms of \( x \) using the volume constraint?
3. What would change in the formula if the box had a closed top?
4. How does the volume constraint influence the shape of the box?
5. Why might minimizing surface area be important in a practical setting?

**Tip:** Remember that minimizing surface area is key in reducing material costs for production, making this method valuable in real-world applications.

A box with a square base and open top must have a volume of 78732 cm³. Find the dimensions of the box that minimize the amount of material used. Provide a formula for the surface area of the box in terms of x, the length of one side of the square base.

Solve an optimization problem involving a box with a square base and open top, given a fixed volume of 78732 cm³. Learn how to express the surface area function in terms of one variable and use calculus to find the optimal dimensions that minimize material use. Suitable for advanced high school or college students.

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