A box with no top will be made by cutting square of equal size out of the corners of a 33 inch by 58 in rectangular piece of cardboard then folding the side flaps up. Find the maximum volume of such a box

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.

A box with no top will be made by cutting squares of equal size out of the corners of a 33 inch by 58 inch 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 33 inch by 58 inch cardboard. Learn the step-by-step process involving optimization, algebraic functions, and derivatives. Ideal for students in Grades 10-12.

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