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To solve this problem, let’s break it down into steps.

### Problem Analysis
We have a box with:
1. A square base.
2. An open top.
3. A required volume of \(2500 \ \text{cm}^3\).

We want to find the dimensions that minimize the surface area of this box.

### Definitions and Variables
Let:
- \( x \) be the length of a side of the square base (in cm).
- \( h \) be the height of the box (in cm).

The volume \( V \) of the box is given by:
\[
V = x^2 \cdot h
\]

Since \( V = 2500 \ \text{cm}^3 \), we have:
\[
x^2 \cdot h = 2500
\]
Solving for \( h \):
\[
h = \frac{2500}{x^2}
\]

### Surface Area Calculation
Since the box has an open top, the surface area \( S \) consists of the area of the base and the area of the four sides:
\[
S = x^2 + 4 \cdot (x \cdot h)
\]
Substitute \( h = \frac{2500}{x^2} \) into this equation:
\[
S = x^2 + 4 \cdot \left(x \cdot \frac{2500}{x^2}\right)
\]
Simplify the equation:
\[
S = x^2 + \frac{10000}{x}
\]

### Minimizing the Surface Area
To find the minimum surface area, take the derivative of \( S \) with respect to \( x \) and set it to zero:
1. Differentiate \( S \) with respect to \( x \):
   \[
   \frac{dS}{dx} = 2x - \frac{10000}{x^2}
   \]
2. Set the derivative to zero:
   \[
   2x - \frac{10000}{x^2} = 0
   \]
   Rearrange to solve for \( x \):
   \[
   2x^3 = 10000
   \]
   \[
   x^3 = 5000
   \]
   \[
   x = \sqrt[3]{5000} \approx 17.1
   \]

### Solving for \( h \)
Using \( x \approx 17.1 \) cm in \( h = \frac{2500}{x^2} \):
\[
h = \frac{2500}{(17.1)^2} \approx 8.55
\]

### Calculating the Minimum Surface Area
Now, substitute \( x \approx 17.1 \) and \( h \approx 8.55 \) back into the surface area formula:
\[
S = (17.1)^2 + \frac{10000}{17.1} \approx 292 + 585 \approx 877 \ \text{cm}^2
\]

### Answer
The minimum possible surface area, rounded to the nearest whole number, is:
\[
\boxed{877}
\]

Would you like a detailed breakdown of any step or further questions?

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#### Further Related Questions
1. How do we determine if this value is a minimum using the second derivative test?
2. What would the surface area be if the box had a closed top?
3. How does changing the volume affect the minimum surface area?
4. Can we use calculus techniques to minimize other constraints, such as perimeter?
5. How does the cube root calculation affect the accuracy of the result?

#### Tip
To minimize problems involving volume and surface area, try expressing one variable in terms of the other using given constraints. Then, apply calculus to find extreme values.

A box with a square base and open top must have a volume of 2500 cm^3. What is the minimum possible surface area (in cm^2) of this box? Round to the nearest whole number and omit units.

Find the minimum possible surface area for a box with a square base and open top, given a fixed volume of 2500 cm^3. This problem explores concepts in calculus and optimization, including volume, surface area, and derivative-based minimization. Suitable for high school students in Grades 11-12.

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