91Ó°ÊÓ

When working with rectangular pieces of cardboard, understanding the initial dimensions is essential. In this exercise, we start with a piece of cardboard where the length is twice the width. This means if the width is measured at, say, 4 feet, the length will automatically be 8 feet because it is calculated as double the width.
Once the dimensions are established, the next step involves altering the cardboard. In this case, we cut a 1-foot by 1-foot square from each of the four corners. This allows us to create flaps that can be folded up to form the sides of the box. By removing these squares, the dimensions of the box's base decrease. So, the potential new length becomes the original length minus 2 feet (1 foot cut from each end), and the width becomes the original width minus 2 feet.
Understanding how these cuts change the cardboard's dimensions is crucial for calculating the resulting box dimensions accurately.

Quadratic equations are powerful tools for solving problems involving dimensions, particularly when those problems aren't straightforward. When we try to create a box from a piece of cardboard by cutting and folding, we often end up simplifying the problem into a quadratic equation to solve for unknown dimensions.
The quadratic equation in this context helps us determine the original width of the cardboard. The expression \[ (2w-2)(w-2) = 12 \] comes from setting up the volume equation after cutting out the corners. Expanding this results in a quadratic equation:\[ 2w^2 - 6w + 4 = 12 \].
After simplifying and rearranging, we get:\[ w^2 - 3w - 4 = 0 \].
The solutions to this equation, typically achieved by factoring or using the quadratic formula, represent the potential values for the width. In our example, only positive solutions make sense, ultimately giving us the width as 4 feet.

Calculating the volume of a three-dimensional object like a box requires knowing its length, width, and height. In the context of our cardboard box, once we've altered the cardboard, we need to ensure it achieves the right volume, which is specified as 12 cubic feet.
The formula for volume is straightforward: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \].
In this problem, once the squares are cut out and the sides are folded, the box's dimensions are adjusted to \[ 6 \times 2 \times 1 \] feet. Here, 6 feet is the new length, 2 feet is the new width, and 1 foot is the height. Multiplying these gives the desired volume of 12 cubic feet.
This formulation confirms our adjustment was correct, as maintaining these dimensional relationships ensures the box achieves the necessary size to meet the specified volume.