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.

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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).