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.