To find the maximum volume of the open-top box made from the cardboard, we'll first define the variables and formulate the problem mathematically.

1. Let $x$ be the side length of the square cut from each corner.

2. After cutting out these squares, the dimensions of the base of the box will be: $\text{Length} = 58 - 2x$ $\text{Width} = 33 - 2x$

3. The height of the box will be $x$.

4. The volume $V$ of the box can be expressed as: $V = \text{Length} \times \text{Width} \times \text{Height} = (58 - 2x)(33 - 2x)x$

5. Volume Function: $V = (58 - 2x)(33 - 2x)x = 4x^3 - 182x^2 + 1914x$

6. Finding the Maximum Volume: To find the maximum volume, we need to find the critical points by taking the derivative of $V$ with respect to $x$ and setting it equal to zero:

   $\frac{dV}{dx} = 12x^2 - 364x + 1914 = 0$

   Solving this quadratic equation will give us the values of $x$ at which the volume could be maximized.

7. Calculating Critical Points: Let's solve this equation to find the critical points.The critical points for the volume function are:

$x = \frac{91 - \sqrt{2539}}{6} \approx 5.96$ $x = \frac{91 + \sqrt{2539}}{6} \approx 26.04$

Given the constraints of the box, $x$ must be less than half of the smaller dimension of the cardboard, which is $\frac{33}{2} = 16.5$. Therefore, only the first critical point $x \approx 5.96$ is within the acceptable range.

8. Calculating Maximum Volume: Let's compute the volume at $x \approx 5.96$.The maximum volume of the box is approximately $5857.35$ cubic inches when the square cut-out size is about $6.77$ inches.

Would you like further details on this solution or have any specific questions?

Related Questions:

1. How do you determine if a critical point is a maximum or minimum?
2. What are the constraints on the value of $x$ for this problem?
3. How would the solution change if the cardboard had a different dimension?
4. Can you show the complete calculation for the critical point derivation?
5. What happens to the volume if you cut squares larger than $16.5$ inches?

Tip: Always verify that the critical points fall within the physical constraints of the problem before concluding they are maxima or minima.