AP Calculus AB 5.11 Optimization: The Open Box-Top Word Problem

**AP Calculus AB 5.11 Optimization: The Open Box-Top Word Problem**

This section focuses on solving optimization problems involving the construction of an open-top box from a rectangular sheet of material. It highlights the application of calculus to find the dimensions that maximize the volume of the box, given specific constraints.

### Problem Setup

1. **Understanding the Scenario**:
- You start with a rectangular sheet of material and want to create an open-top box by cutting out squares from each corner and folding up the sides.
- The size of the squares cut from each corner will affect both the dimensions of the base of the box and its height.

2. **Defining Variables**:
- Let \( L \) represent the length of the original sheet, \( W \) represent the width, and \( x \) represent the side length of the squares cut from each corner.
- The new dimensions of the base of the box will be \( (L - 2x) \) for the length and \( (W - 2x) \) for the width.
- The height of the box will be \( x \).

3. **Volume Function**:
- The volume \( V \) of the box can be expressed as:
\[
V = \text{length} \times \text{width} \times \text{height} = (L - 2x)(W - 2x)x
\]
- This function can be expanded to:
\[
V = x(LW - 2Lx - 2Wx + 4x^2)
\]

### Optimization Process

1. **Finding the Critical Points**:
- To maximize the volume, take the derivative of the volume function \( V \) with respect to \( x \):
\[
V' = \frac{dV}{dx}
\]
- Set the derivative equal to zero to find critical points:
\[
V' = 0
\]
- Solving this equation will give potential values for \( x \).

2. **Constraints**:
- It's essential to consider the constraints of the problem. The value of \( x \) must be greater than zero and less than \( \frac{L}{2} \) and \( \frac{W}{2} \) to ensure that the cuts do not exceed the dimensions of the original sheet.

3. **Second Derivative Test**:
- To confirm that the critical point corresponds to a maximum volume, take the second derivative of the volume function:
\[
V'' = \frac{d^2V}{dx^2}
\]
- If the second derivative is negative at the critical point, it indicates that the function is concave down and that the critical point represents a maximum.

### Conclusion

The optimal size of the squares cut from each corner can be found by analyzing the critical points within the constraints of the problem. This process illustrates how to apply calculus concepts to solve real-world problems and emphasizes the importance of reasoning through both mathematical relationships and physical constraints.

### Summary

This problem serves as a practical application of calculus to optimize dimensions in a real-world scenario. By setting up a volume function based on defined variables, differentiating to find critical points, and verifying those points with the second derivative test, you can find the dimensions that yield the maximum volume for the open-top box.

**Key Takeaways**:
- Use calculus to find maximum or minimum values for functions describing real-world scenarios.
- Define variables carefully and consider constraints in optimization problems.
- The open-top box problem exemplifies how to use derivatives and critical points in practical applications, reinforcing key calculus concepts.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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