Lorene plans to make several open-topped boxes in which to carry plants. The boxes are made from rectangular sheets of cardboard from which Lorene cuts out
8-in squares from each corner. The length of the original piece of cardboard is 16 in more than the width. If the volume of the box is 136 in3 , determine the dimensions of the original piece of cardboard.

To solve this problem, we need to find the dimensions of the original piece of cardboard, given that Lorene cuts out 8-inch squares from each corner to form open-topped boxes, and the volume of each box is 136 cubic inches.

### Step 1: Define the dimensions of the original cardboard.
Let:
- \( w \) be the width of the original rectangular piece of cardboard (in inches).
- \( l \) be the length of the original rectangular piece of cardboard. Since the length is 16 inches more than the width, we have:
  \[
  l = w + 16
  \]

### Step 2: Dimensions of the box after cutting.
After cutting 8-inch squares from each corner, the dimensions of the base of the box will be:
- Width of the box: \( w - 2 \times 8 = w - 16 \)
- Length of the box: \( (w + 16) - 2 \times 8 = w \)
- Height of the box: 8 inches (since the squares cut out are 8 inches).

### Step 3: Set up the volume equation.
The volume \( V \) of a rectangular box is given by:
\[
V = \text{length} \times \text{width} \times \text{height}
\]
Substituting the values:
\[
136 = w \times (w - 16) \times 8
\]
Now, divide both sides of the equation by 8:
\[
17 = w \times (w - 16)
\]

### Step 4: Solve the quadratic equation.
Expand the equation:
\[
17 = w^2 - 16w
\]
Rearrange it to form a standard quadratic equation:
\[
w^2 - 16w - 17 = 0
\]
Solve this using the quadratic formula:
\[
w = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(-17)}}{2(1)}
\]
\[
w = \frac{16 \pm \sqrt{256 + 68}}{2}
\]
\[
w = \frac{16 \pm \sqrt{324}}{2}
\]
\[
w = \frac{16 \pm 18}{2}
\]
So we have two possible solutions:
\[
w = \frac{16 + 18}{2} = 17 \quad \text{or} \quad w = \frac{16 - 18}{2} = -1
\]
Since the width cannot be negative, we take \( w = 17 \).

### Step 5: Find the length.
Since \( l = w + 16 \), the length is:
\[
l = 17 + 16 = 33 \, \text{inches}
\]

### Step 6: Verify the solution.
The dimensions of the box are:
- Width of the box: \( 17 - 16 = 1 \, \text{inch} \)
- Length of the box: \( 17 \, \text{inches} \)
- Height of the box: \( 8 \, \text{inches} \)

Now, compute the volume:
\[
V = 17 \times 1 \times 8 = 136 \, \text{in}^3
\]
This matches the given volume, so the solution is correct.

### Final Answer:
The dimensions of the original piece of cardboard are:
- Width: 17 inches
- Length: 33 inches

Would you like further details on any of these steps or have any other questions?

Here are 5 related questions you might find interesting:
1. How would the problem change if the squares cut out were 10 inches instead of 8 inches?
2. How does the volume formula change if the box has a lid?
3. What would the dimensions be if the volume was 200 cubic inches instead of 136?
4. Can this method be applied if the length and width relationship is different, such as the length being twice the width?
5. How can the quadratic formula be derived from first principles?

**Tip:** When solving word problems, always start by clearly defining the variables and relationships before setting up equations. This helps prevent confusion later on.

Lorene plans to make several open-topped boxes in which to carry plants. The boxes are made from rectangular sheets of cardboard from which Lorene cuts out 8-in squares from each corner. The length of the original piece of cardboard is 16 in more than the width. If the volume of the box is 136 in³, determine the dimensions of the original piece of cardboard.

Learn how to determine the dimensions of a cardboard sheet used to create an open-topped box by solving a quadratic equation. This problem involves concepts of algebra and geometry, using the volume formula for a rectangular box. Suitable for students in Grades 9-11.

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