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To solve the problem of maximizing the volume of the open-top box created from a 16" by 14" piece of metal by cutting out squares from each corner and folding up the sides, we can follow these steps:

1. **Define Variables:**
   - Let \( x \) be the side length of the squares cut from each corner.

2. **Dimensions of the Box:**
   - After cutting out squares with side \( x \) from each corner, the new dimensions of the base will be \( 16 - 2x \) and \( 14 - 2x \).
   - The height of the box will be \( x \).

3. **Volume of the Box:**
   - The volume \( V \) of the box can be expressed as:
     \[
     V = x (16 - 2x)(14 - 2x)
     \]

4. **Formulate the Volume Function:**
   \[
   V(x) = x (16 - 2x)(14 - 2x)
   \]

5. **Expand the Volume Function:**
   \[
   V(x) = x (224 - 32x - 28x + 4x^2)
   \]
   \[
   V(x) = x (224 - 60x + 4x^2)
   \]
   \[
   V(x) = 224x - 60x^2 + 4x^3
   \]

6. **Find the Critical Points:**
   - To find the maximum volume, we take the derivative of \( V(x) \) and set it to zero:
     \[
     V'(x) = 224 - 120x + 12x^2
     \]
   - Set \( V'(x) = 0 \):
     \[
     12x^2 - 120x + 224 = 0
     \]
   - Simplify the quadratic equation:
     \[
     3x^2 - 30x + 56 = 0
     \]
   - Use the quadratic formula to find \( x \):
     \[
     x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
     \]
     where \( a = 3 \), \( b = -30 \), and \( c = 56 \).

7. **Solve the Quadratic Equation:**
   \[
   x = \frac{30 \pm \sqrt{900 - 672}}{6}
   \]
   \[
   x = \frac{30 \pm \sqrt{228}}{6}
   \]
   \[
   x = \frac{30 \pm 2\sqrt{57}}{6}
   \]
   \[
   x = 5 \pm \frac{\sqrt{57}}{3}
   \]

   - Since \( x \) must be positive and smaller than half the smallest side of the metal piece:
     \[
     x = 5 - \frac{\sqrt{57}}{3}
     \]

8. **Calculate the Maximum Volume:**
   - Substitute \( x \) back into the volume formula to find the maximum volume.

Do you want me to continue and find the exact maximum volume or any further details on this calculation?

### Related Questions:

1. How would the solution change if the metal piece was a different size?
2. What would be the effect of changing the shape of the cut-out corners (e.g., circles instead of squares)?
3. How does the thickness of the metal affect the box's maximum volume?
4. Can this method be used to optimize other types of box shapes (e.g., with lids)?
5. How does the calculated optimal height compare to practical considerations for folding metal?

### Tip:
When dealing with optimization problems involving dimensions, always verify the physical feasibility of your solution within the given constraints.

An open-top box is being designed by cutting a corner piece out of a 16" by 14" piece of metal and folding the sides upwards. The designer wants to maximize the volume of this box.

Learn how to maximize the volume of an open-top box created from a 16-inch by 14-inch piece of metal. This problem involves quadratic equations, derivative calculations, and optimization techniques, suitable for Grades 10-12.

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