The image shows a performance task related to constructing a box without a cover based on a given procedure. The steps involve taking a rectangle of dimensions 6 units by 8 units, cutting out equal squares from each corner, and folding the remaining flaps up to form the sides of a box without a cover.

To solve this:

1. Step 1: Start with a rectangle that measures 6 units by 8 units.

2. Step 2: Squares are cut from each corner. Let the side length of each square be x. The remaining central section will form the base, while the folded-up flaps will form the sides of the box.

   The new dimensions of the base will be:

   • Length: $6 - 2x$ units.
   • Width: $8 - 2x$ units.
   • Height of the box: $x$ units (from the squares cut out).

3. Volume of the resulting box: The volume $V$ of a box is calculated using the formula for the volume of a rectangular prism: $V = \text{length} \times \text{width} \times \text{height}$ Substituting the dimensions: $V = (6 - 2x)(8 - 2x)(x)$

To find the volume, you'll need to know the value of $x$, which can be a variable choice depending on how much is cut out.

Insights from this activity:

• The volume depends on how much material is cut from the corners.
• There's an optimal value for $x$ that maximizes the volume.
• The problem is a great practical application of algebraic expressions and volume formulas.

Would you like further details on how to maximize the volume, or any clarification on the mathematical steps?

Questions for further thought:

1. What value of $x$ maximizes the volume of the box?
2. How does changing $x$ affect the box's volume?
3. What happens if no squares are cut out (i.e., $x = 0$)?
4. How could you find the optimal square size $x$ mathematically?
5. Could this method be applied to other shapes or sizes of paper?

Tip:

When working on optimization problems involving volume, remember to check for constraints and use calculus (derivatives) if needed to find the maximum value.