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The image depicts a problem involving an open-top box with a length of \( 52 - 2x \) inches, a width of \( 42 - 2x \) inches, and a height of \( x \) inches. The problem requires solving several parts:

### A. **Describing how to make the box from a rectangular piece of cardboard**:
To create the box, start with a rectangular piece of cardboard with original dimensions \( 52 \) inches by \( 42 \) inches. Cut out squares of side \( x \) inches from each corner of the cardboard. After removing these squares, fold up the flaps created by cutting the squares to form the sides of the box. This creates the open-top box with dimensions \( (52 - 2x) \times (42 - 2x) \times x \).

### B. **Degree of the polynomial representing the volume**:
The volume \( V(x) \) of the box is calculated by multiplying the length, width, and height:
\[
V(x) = (52 - 2x)(42 - 2x)(x)
\]
First, expand the product of the length and width:
\[
(52 - 2x)(42 - 2x) = 2184 - 188x + 4x^2
\]
Now, multiply by the height \( x \):
\[
V(x) = x(2184 - 188x + 4x^2) = 2184x - 188x^2 + 4x^3
\]
Thus, the volume is represented by the polynomial:
\[
V(x) = 4x^3 - 188x^2 + 2184x
\]
The degree of this polynomial is **3** because the highest power of \( x \) is \( x^3 \).

### C. **Procedure for finding the value of \( x \) that maximizes the volume**:
To find the value of \( x \) that maximizes the volume, we need to find the critical points by taking the derivative of \( V(x) \), setting it equal to zero, and solving for \( x \).

The derivative of the volume function is:
\[
V'(x) = 12x^2 - 376x + 2184
\]
Set this equal to zero to find the critical points:
\[
12x^2 - 376x + 2184 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-376) \pm \sqrt{(-376)^2 - 4(12)(2184)}}{2(12)}
\]
This will give the value of \( x \) that maximizes the volume.

Let me know if you would like me to continue solving this quadratic equation!

---

Would you like further details or have any questions?

Here are 5 related questions:
1. What are the possible values of \( x \) based on the dimensions of the box?
2. Why is the degree of the volume polynomial important?
3. How would you interpret the critical points in the context of maximizing volume?
4. Could there be a minimum volume for certain values of \( x \), or is it always positive?
5. How does cutting squares from the cardboard influence the overall volume?

**Tip**: When solving optimization problems, always verify that the solution is within the problem's domain (i.e., check that the dimensions make sense physically).

As shown, an open box has a length of (52-2x) inches, a width of (42-2x) inches, and a height of x inches. Describe how to create this box from a rectangular piece of cardboard, identify the degree of the polynomial that represents the volume, and describe a procedure for finding the value of x that yields the maximum possible volume.

This problem explores how to create an open-top box from a piece of cardboard by cutting squares from the corners. It involves finding the volume polynomial and using derivatives to determine the value of x that maximizes the volume. Suitable for students in Grades 10-12.

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