A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal square are removed from the corners. The sides of this sheet are now turned up to form an open rectangular box. Calculate the height (in cm) of the box when the volume is maximum.

To solve this problem, we need to determine the height of the open rectangular box that maximizes its volume after cutting equal squares from each corner of a rectangular sheet.

### Step-by-Step Solution

1. **Understand the Problem Setup:**
   - The original sheet has a length of \(6\) meters and a width of \(2\) meters.
   - We cut out four equal squares from each corner of the sheet. Let the side length of each square be \(x\) meters.
   - After cutting out the squares, the sheet is folded to form an open rectangular box.

2. **Dimensions of the Box:**
   - After cutting out squares of side \(x\) meters from each corner, the length and width of the base of the box become:
     \[
     \text{New length} = 6 - 2x \quad \text{(since \(x\) is cut from both sides)}
     \]
     \[
     \text{New width} = 2 - 2x \quad \text{(since \(x\) is cut from both sides)}
     \]
   - The height of the box is \(x\) meters.

3. **Volume of the Box:**
   - The volume \(V\) of the box can be expressed as:
     \[
     V = \text{length} \times \text{width} \times \text{height}
     \]
     Substituting the expressions for the length, width, and height:
     \[
     V(x) = (6 - 2x)(2 - 2x)(x)
     \]

4. **Simplify the Volume Formula:**
   \[
   V(x) = x(6 - 2x)(2 - 2x)
   \]
   \[
   V(x) = x(12 - 12x + 4x^2)
   \]
   \[
   V(x) = 12x - 12x^2 + 4x^3
   \]

5. **Find the Value of \(x\) that Maximizes the Volume:**
   - To maximize the volume, we take the derivative of \(V(x)\) with respect to \(x\) and set it to zero:
   \[
   V'(x) = 12 - 24x + 12x^2
   \]
   - Set \(V'(x) = 0\):
   \[
   12 - 24x + 12x^2 = 0
   \]
   - Divide the equation by 12:
   \[
   1 - 2x + x^2 = 0
   \]
   - Rearrange to get a standard quadratic equation:
   \[
   x^2 - 2x + 1 = 0
   \]
   - Factor the quadratic:
   \[
   (x - 1)^2 = 0
   \]
   - Solving for \(x\):
   \[
   x = 1
   \]

6. **Check if \(x = 1\) is a Maximum:**
   - To confirm it's a maximum, we check the second derivative \(V''(x)\):
   \[
   V''(x) = -24 + 24x
   \]
   - Substituting \(x = 1\):
   \[
   V''(1) = -24 + 24 = 0
   \]
   Since \(V''(1) = 0\), we must use the first derivative test around \(x = 1\):
   - Check \(V'(x)\) for \(x < 1\) and \(x > 1\). 
   - If \(x < 1\) (e.g., \(x = 0.5\)):
     \[
     V'(0.5) = 12 - 24(0.5) + 12(0.5)^2 = 12 - 12 + 3 = 3 > 0
     \]
   - If \(x > 1\) (e.g., \(x = 1.5\)):
     \[
     V'(1.5) = 12 - 24(1.5) + 12(1.5)^2 = 12 - 36 + 27 = 3 > 0
     \]

   This confirms that \(V'(x)\) changes from positive to zero around \(x = 1\), indicating a local maximum at \(x = 1\).

7. **Convert the Height to Centimeters:**
   - The optimal height \(x\) of the box in meters is \(1\) meter.
   - Converting to centimeters:
     \[
     1 \text{ meter} = 100 \text{ cm}
     \]

### Conclusion:
The height of the box when the volume is maximum is **100 cm**.

---

Would you like more details on this calculation or have any other questions?

Here are some related questions to explore:

1. How would the volume change if the length of the rectangular sheet was different?
2. What is the general formula for the volume of a box formed from any rectangular sheet?
3. How does the size of the square cutouts affect the volume?
4. Can the dimensions of the sheet affect the feasibility of constructing a box?
5. What other shapes could be formed by manipulating a rectangular sheet?

**Tip:** Always check units when converting between measurements, especially when dealing with different systems (e.g., meters to centimeters).

Learn how to maximize the volume of a rectangular box formed by cutting corners from a metal sheet. Step-by-step solution provided with calculations and optimal height determination. Suitable for advanced high school students and beyond.

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