Let's solve the problem based on the given information.

Part a:

The height of the box is represented by $x$, as it’s the height created by cutting out squares with side length $x$ and folding up the sides. Therefore:

$\text{Height of the box} = x \text{ inches}$

Part b:

To find the volume of the box, we use the formula for the volume of a rectangular prism:

$\text{Volume} = \text{length} \times \text{width} \times \text{height}$

After cutting out squares of $x$ inches from each corner, the dimensions of the box will be as follows:

• New length = $16 - 2x$ (since $x$ inches are cut from both ends of the original 16-inch length)
• New width = $11 - 2x$ (since $x$ inches are cut from both ends of the original 11-inch width)
• Height = $x$

Thus, the volume $V$ of the box is:

$V = (16 - 2x)(11 - 2x)(x)$

Simplifying the Expression for Volume

1. Expand $(16 - 2x)(11 - 2x)$:

   $(16 - 2x)(11 - 2x) = 176 - 22x - 32x + 4x^2 = 176 - 54x + 4x^2$

2. Now, multiply by $x$:

   $V = x(176 - 54x + 4x^2)$

   Distribute $x$:

   $V = 4x^3 - 54x^2 + 176x$

So, the simplified expression for the volume of the box is:

$V = 4x^3 - 54x^2 + 176x \text{ cubic inches}$

Would you like further clarification on any step?

Here are some questions to further explore the topic:

1. What happens to the volume if the value of $x$ is too large?
2. How would the formula change if the original dimensions were different?
3. Can we calculate the maximum possible volume for this box using calculus?
4. What are the practical limitations for choosing $x$ based on the cardboard dimensions?
5. How does this problem relate to polynomial functions and their applications?

Tip: When simplifying expressions, expand terms carefully and combine like terms to avoid mistakes.