A box will be made by cutting squares of equal size out of the corners of a 40” x 53 interacting with a piece of cardboard and fold inside flips up find the maximum volume of such a box

To find the maximum volume of the box that can be formed by cutting squares from the corners of the 40-inch by 53-inch cardboard, let's proceed step by step.

### Step 1: Define variables
Let \( x \) be the side length of the square cut from each corner. After cutting out these squares and folding up the flaps, the dimensions of the resulting box will be:

- Length = \( 53 - 2x \)
- Width = \( 40 - 2x \)
- Height = \( x \)

### Step 2: Express the volume of the box
The volume \( V \) of the box can be expressed as:
\[
V(x) = (53 - 2x)(40 - 2x)(x)
\]
This represents the length, width, and height of the box after folding.

### Step 3: Expand the volume equation
We can expand this equation to make it easier to work with:
\[
V(x) = x(53 - 2x)(40 - 2x)
\]
First, expand the product inside the parentheses:
\[
(53 - 2x)(40 - 2x) = 53 \times 40 - 53 \times 2x - 40 \times 2x + 4x^2
\]
\[
= 2120 - 106x - 80x + 4x^2
\]
\[
= 2120 - 186x + 4x^2
\]
Now multiply this by \( x \):
\[
V(x) = x(2120 - 186x + 4x^2)
\]
\[
V(x) = 2120x - 186x^2 + 4x^3
\]
This is the volume function we need to maximize.

### Step 4: Find the critical points
To find the maximum volume, take the derivative of \( V(x) \) with respect to \( x \) and set it equal to zero:
\[
V'(x) = 2120 - 2 \times 186x + 3 \times 4x^2
\]
\[
V'(x) = 2120 - 372x + 12x^2
\]
Set \( V'(x) = 0 \) to find the critical points:
\[
12x^2 - 372x + 2120 = 0
\]
### Step 5: Solve the quadratic equation
Use the quadratic formula to solve for \( x \):
\[
x = \frac{-(-372) \pm \sqrt{(-372)^2 - 4(12)(2120)}}{2(12)}
\]
\[
x = \frac{372 \pm \sqrt{138384 - 101760}}{24}
\]
\[
x = \frac{372 \pm \sqrt{36624}}{24}
\]
\[
x = \frac{372 \pm 191.36}{24}
\]
This gives two solutions for \( x \):
\[
x_1 = \frac{372 + 191.36}{24} \approx 23.53 \quad \text{and} \quad x_2 = \frac{372 - 191.36}{24} \approx 7.53
\]

### Step 6: Analyze the critical points
Since \( x = 23.53 \) is too large for the dimensions of the cardboard, we discard it. The valid solution is \( x = 7.53 \).

### Step 7: Calculate the maximum volume
Substitute \( x = 7.53 \) into the volume equation:
\[
V(7.53) = 2120(7.53) - 186(7.53)^2 + 4(7.53)^3
\]
Let's compute this.

The maximum volume of the box is approximately 7125.06 cubic inches when squares of side length 7.53 inches are cut from the corners of the 40" x 53" cardboard.

Would you like more details or have any further questions? 

Here are 5 related questions for further exploration:
1. What happens if the dimensions of the cardboard are changed?
2. How does the size of the cut squares affect the volume?
3. Can we maximize the volume for different shapes of cardboard, like a square piece?
4. How would the process change if the cardboard had a thickness?
5. What is the surface area of the resulting box?

**Tip**: The derivative of a function helps find its maximum or minimum, which is crucial in optimization problems.

A box will be made by cutting squares of equal size out of the corners of a 40” x 53” piece of cardboard and folding the flaps up. Find the maximum volume of such a box.

Learn how to maximize the volume of a box created by cutting squares from the corners of a 40" x 53" cardboard sheet. This problem involves key concepts of optimization, quadratic equations, and volume calculation. Suitable for high school students in Grades 10-12.

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