https://file.aiquickdraw.com/mathbot/file/1725486292414efwwnyce.jpg

Let's analyze the problem step by step.

### Problem Breakdown:
1. **Dimensions of the Cardboard:**
   - Length: 48 cm
   - Width: 30 cm
   - Squares of side length \(x\) cm are cut from each corner.

2. **Resulting Box Dimensions:**
   After cutting the squares and folding, the dimensions of the box will be:
   - Length: \(48 - 2x\) cm
   - Width: \(30 - 2x\) cm
   - Height: \(x\) cm

3. **Volume Function:**
   The volume \(V(x)\) of the box as a function of \(x\) is given by:
   \[
   V(x) = (48 - 2x)(30 - 2x)(x)
   \]
   for \(0 < x < 15\), because the maximum side length of the square cutout cannot exceed half of the smaller dimension (30 cm).

### Questions in the Problem:
(a) Write down the maximum possible value of \(V(x)\) in its context.
(b) Find the value of \(x\) that maximizes the volume of the box.

### Step-by-Step Solution:

1. **Volume Function:**
   Expanding the volume function:
   \[
   V(x) = (48 - 2x)(30 - 2x)(x)
   \]
   First, expand the first two terms:
   \[
   (48 - 2x)(30 - 2x) = 1440 - 96x - 60x + 4x^2 = 1440 - 156x + 4x^2
   \]
   Now, multiply by \(x\):
   \[
   V(x) = x(1440 - 156x + 4x^2) = 4x^3 - 156x^2 + 1440x
   \]
   So the volume function becomes:
   \[
   V(x) = 4x^3 - 156x^2 + 1440x
   \]

2. **Find Maximum Volume:**
   To find the maximum volume, we need to take the derivative of \(V(x)\) and find the critical points.
   \[
   V'(x) = 12x^2 - 312x + 1440
   \]
   Setting \(V'(x) = 0\) to find the critical points:
   \[
   12x^2 - 312x + 1440 = 0
   \]
   Dividing by 12:
   \[
   x^2 - 26x + 120 = 0
   \]
   Solving this quadratic equation using the quadratic formula:
   \[
   x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(1)(120)}}{2(1)}
   \]
   Simplifying:
   \[
   x = \frac{26 \pm \sqrt{676 - 480}}{2} = \frac{26 \pm \sqrt{196}}{2} = \frac{26 \pm 14}{2}
   \]
   So the roots are:
   \[
   x = \frac{26 + 14}{2} = 20 \quad \text{and} \quad x = \frac{26 - 14}{2} = 6
   \]
   Since \(0 < x < 15\), the valid critical point is \(x = 6\).

3. **Maximum Volume:**
   To confirm \(x = 6\) gives a maximum, we can evaluate the second derivative or compare values at endpoints. Substituting \(x = 6\) into the volume function:
   \[
   V(6) = 4(6)^3 - 156(6)^2 + 1440(6) = 4(216) - 156(36) + 8640 = 864 - 5616 + 8640 = 3888 \, \text{cm}^3
   \]
   Thus, the maximum possible volume is \(3888 \, \text{cm}^3\) when \(x = 6\) cm.

### Final Answers:
- (a) The maximum possible value of \(V(x)\) is \(3888 \, \text{cm}^3\).
- (b) The value of \(x\) that maximizes the volume is \(x = 6\) cm.

Would you like any additional details or further explanations? 

Here are five related questions:
1. How would the problem change if the cardboard had different dimensions?
2. Can you derive the volume function if the cardboard dimensions were reversed (e.g., 30 cm by 48 cm)?
3. What would be the impact on the volume if the squares cut from the corners were different sizes?
4. How can you verify that \(x = 6\) gives a maximum using the second derivative test?
5. What if the box had a closed top? How would the volume function change?

**Tip:** To solve cubic volume problems efficiently, always expand the equation and simplify as much as possible before applying calculus techniques.

A rectangular box with an open top is constructed from a cardboard piece measuring 48 cm by 30 cm. Squares of equal size are cut from the corners, and the box is folded. The volume is modeled by V(x) = (48 - 2x)(30 - 2x)(x), where x is the side length of the squares. (a) Write down the maximum possible value of V(x). (b) Find the value of x that maximizes the volume.

Learn how to solve an optimization problem involving the construction of an open-top box from a cardboard piece. This problem models the volume function and uses calculus techniques to find the maximum possible volume. Suitable for students in Grades 11-12.

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