A box with no top will be made by cutting squares of equalize out of the corners of a 36” x 51” rectangular piece of cardboard then folding the side flaps up. Find the maximum volume of such a box.

The problem is asking to find the maximum volume of an open-top box created by cutting squares from the corners of a rectangular piece of cardboard and then folding up the sides.

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

After cutting the squares, the dimensions of the box will be:
- **Length**: \( 51 - 2x \) (since two squares are cut, one from each side of the length)
- **Width**: \( 36 - 2x \) (similarly for the width)
- **Height**: \( x \) (because the side flaps are folded up to form the height of the box)

### Step 2: Volume of the Box
The volume \( V \) of the box is given by:
\[
V(x) = \text{Length} \times \text{Width} \times \text{Height} = (51 - 2x)(36 - 2x)(x)
\]
Simplify this expression:
\[
V(x) = x(51 - 2x)(36 - 2x)
\]

### Step 3: Expand the Expression
First, expand the product \( (51 - 2x)(36 - 2x) \):
\[
(51 - 2x)(36 - 2x) = 51 \times 36 - 51 \times 2x - 2x \times 36 + 4x^2
\]
\[
= 1836 - 102x - 72x + 4x^2
\]
\[
= 1836 - 174x + 4x^2
\]

Now, multiply this result by \( x \):
\[
V(x) = x(1836 - 174x + 4x^2) = 1836x - 174x^2 + 4x^3
\]

Thus, the volume function is:
\[
V(x) = 4x^3 - 174x^2 + 1836x
\]

### Step 4: Maximize the Volume
To find the value of \( x \) that maximizes the volume, take the derivative of \( V(x) \) with respect to \( x \), and set it equal to zero:
\[
V'(x) = 12x^2 - 348x + 1836
\]
Set \( V'(x) = 0 \) to find the critical points:
\[
12x^2 - 348x + 1836 = 0
\]
This is a quadratic equation. We can solve it using the quadratic formula:
\[
x = \frac{-(-348) \pm \sqrt{(-348)^2 - 4(12)(1836)}}{2(12)}
\]
\[
x = \frac{348 \pm \sqrt{121104 - 88128}}{24}
\]
\[
x = \frac{348 \pm \sqrt{32976}}{24}
\]
\[
x = \frac{348 \pm 181.59}{24}
\]

Thus, the two solutions are:
\[
x = \frac{348 + 181.59}{24} \approx 22.04 \quad \text{or} \quad x = \frac{348 - 181.59}{24} \approx 6.94
\]

### Step 5: Verify the Maximum
Since \( x = 22.04 \) is too large (it would make the dimensions negative), we consider \( x = 6.94 \).

### Step 6: Calculate the Maximum Volume
Substitute \( x = 6.94 \) into the volume function:
\[
V(6.94) = 4(6.94)^3 - 174(6.94)^2 + 1836(6.94)
\]
Let me calculate this value.The maximum volume of the box is approximately \( 5698.4 \) cubic inches when \( x = 6.94 \) inches.

Would you like more details or have any questions?

Here are some related questions to expand the topic:
1. How does the size of the cut squares affect the volume of the box?
2. What happens to the volume if the rectangular cardboard is a different size?
3. Can you create a formula to find the optimal \( x \) for any rectangular piece of cardboard?
4. What if the box has a top? How would that change the problem?
5. How would changing the material thickness affect the final box dimensions?

**Tip:** When optimizing, always verify that the dimensions remain physically meaningful (positive) after calculating critical points.

A box with no top will be made by cutting squares of equal size out of the corners of a 36” x 51” rectangular piece of cardboard then folding the side flaps up. Find the maximum volume of such a box.

Find the maximum volume of an open-top box made by cutting squares from the corners of a 36” x 51” rectangular cardboard sheet. Learn how to apply algebra, optimization, and calculus to solve the problem, including using the quadratic formula and derivative tests.

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