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The problem asks to find the maximum volume of a box with no top, which is formed by cutting squares out of the corners of a rectangular cardboard and folding up the sides. Here's the breakdown of how to solve it:

### Step 1: Define variables
Let \( x \) represent the side length of the square cut from each corner.

- The dimensions of the original rectangular piece of cardboard are 25 inches by 48 inches.
- After cutting out squares from the corners, the new dimensions of the base of the box will be:
  - Length = \( 48 - 2x \)
  - Width = \( 25 - 2x \)
  - Height = \( x \) (the height of the box comes from the square cut and folding the sides up).

### Step 2: Volume formula
The volume \( V \) of the box is given by the formula:
\[
V(x) = \text{length} \times \text{width} \times \text{height} = (48 - 2x)(25 - 2x)(x)
\]
This simplifies to:
\[
V(x) = x(48 - 2x)(25 - 2x)
\]

### Step 3: Maximizing the volume
To maximize the volume, we need to take the derivative of \( V(x) \) with respect to \( x \), set it equal to 0, and solve for \( x \). Let's first expand the volume expression:

\[
V(x) = x[(48)(25) - 2(48)x - 2(25)x + 4x^2]
\]
\[
V(x) = x[1200 - 96x - 50x + 4x^2]
\]
\[
V(x) = x(1200 - 146x + 4x^2)
\]
\[
V(x) = 1200x - 146x^2 + 4x^3
\]

Now, take the derivative \( V'(x) \):

\[
V'(x) = 1200 - 292x + 12x^2
\]

Set \( V'(x) = 0 \) to find the critical points:

\[
1200 - 292x + 12x^2 = 0
\]

I will now solve this quadratic equation to find the values of \( x \).The critical points are:

\[
x_1 = \frac{73}{6} - \frac{\sqrt{1729}}{6}
\quad \text{and} \quad
x_2 = \frac{73}{6} + \frac{\sqrt{1729}}{6}
\]

We need to evaluate these points to find their approximate values and check which is within the valid range (since the side length \( x \) cannot be too large, as it has to fit within the original dimensions of the cardboard).

Let's compute the approximate values of these critical points.The critical points are approximately:

- \( x_1 \approx 5.24 \)
- \( x_2 \approx 19.10 \)

Since the side length \( x \) must be less than half of the shorter dimension of the cardboard (i.e., less than \( 25/2 = 12.5 \) inches), the valid critical point is \( x_1 \approx 5.24 \).

### Step 4: Calculate the maximum volume
We can now substitute \( x = 5.24 \) back into the volume formula to find the maximum volume. I'll compute that next.The maximum volume of the box is approximately **2855 cubic inches**, when rounded to the nearest cubic inch.

Would you like a detailed explanation of any part of the process or have any questions?

Here are some related questions to expand your understanding:
1. How do you find the derivative of a volume function?
2. Why is it important to check the physical constraints of the problem when solving for \( x \)?
3. What would happen if you chose the other critical point \( x_2 \)?
4. How does the shape of the original cardboard affect the box's final volume?
5. Can this method be applied to other shapes, such as squares or circles, for similar box constructions?

**Tip:** Always check the domain of your problem to ensure solutions are physically valid (in this case, the side length had to be smaller than 12.5 inches).

A box (with no top) will be made by cutting squares of equal size out of the corners of a 25 inch by 48 inch rectangular piece of cardboard, then folding the side flaps up. Find the maximum volume of such a box. Round to the nearest cubic inch.

Learn how to maximize the volume of a box created from a 25x48 inch rectangular piece of cardboard by cutting equal squares from each corner and folding up the sides. This optimization problem involves derivatives and is suitable for high school students in Grades 10-12. Volume is rounded to the nearest cubic inch.

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