To solve this problem, let’s go through each step systematically.

1. Define the Problem: We have a 23-inch by 45-inch piece of cardboard, and we're cutting congruent squares from each corner. Let's call the side length of each square $x$. After cutting, we fold up the sides to form an open-top box.

2. Determine Box Dimensions:

   • After cutting squares of side length $x$ from each corner:
     • The length of the box will be $45 - 2x$.
     • The width of the box will be $23 - 2x$.
     • The height of the box will be $x$.

3. Set Up the Volume Function: The volume $V$ of a box is given by: $V = \text{length} \times \text{width} \times \text{height}$ Substituting in our expressions for length, width, and height: $V(x) = (45 - 2x)(23 - 2x)x$ Expanding this equation will give a polynomial in terms of $x$, which represents the volume as a function of $x$.

4. Maximize the Volume: To find the maximum volume, we need to find the critical points of $V(x)$ by differentiating and setting the derivative equal to zero: $V'(x) = 0$ Then, we’ll evaluate $V(x)$ at those critical points and within the feasible domain for $x$, which is determined by the constraints $0 < x < 11.5$ (since cutting squares larger than half the width or length would result in negative dimensions).

5. Calculate the Optimal Dimensions: Solving for $x$ will give us the optimal height. Substitute this $x$ back into the expressions for length and width to obtain the final dimensions of the box with maximum volume.

6. Compute the Volume: With the optimal $x$, we substitute back into $V(x)$ to find the maximum volume.

Let me calculate this step-by-step.The optimal dimensions of the box with maximum volume are as follows:

• Length: 35.32 inches
• Width: 13.32 inches
• Height: 4.84 inches

The maximum volume of this box is approximately $2277.04$ cubic inches.

Would you like further details on any part of the solution, or have any questions?

Here are some related questions for further exploration:

1. What would happen to the maximum volume if the dimensions of the cardboard changed?
2. How does changing the height $x$ affect the dimensions and volume of the box?
3. Could a different shape (like a square base) achieve a larger volume with the same cardboard?
4. What if the box needed a lid; how would that change the dimensions for maximum volume?
5. Is there a way to generalize this solution for any rectangular piece of cardboard?

Tip: For similar problems, always start by defining variables and writing equations based on the problem’s constraints. This approach simplifies finding optimal values.