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An open box is to be made from a square piece of material 36 cm on a side by cutting equal

squares with a length of x from the corners and turning the sides up.a) Find the function V(x) for the volume.

b) Determine the domain of V(x).

c) Using a table to estimate the range of the dimensions within which the volume is a maximum.

d) Find the dimensions for the max volume and the max volume.

Let's solve each part of the problem step by step:

### Part (a): Find the function \( V(x) \) for the volume.

To form the open box, you cut out squares of side \( x \) from each corner of a 36 cm square sheet of material. After cutting, the resulting box dimensions will be \( (36 - 2x) \) cm by \( (36 - 2x) \) cm in the base, and \( x \) cm in height (since the sides that are turned up form the height of the box).

The volume \( V \) of a box is given by the formula:
\[
V = \text{length} \times \text{width} \times \text{height}
\]
Here, both the length and width are \( (36 - 2x) \) and the height is \( x \), so:
\[
V(x) = (36 - 2x)(36 - 2x)x = (36 - 2x)^2x
\]
Expanding this, we get:
\[
V(x) = (1296 - 144x + 4x^2)x = 4x^3 - 144x^2 + 1296x
\]

### Part (b): Determine the domain of \( V(x) \).

The domain of \( V(x) \) is constrained by the physical dimensions of the material. Each side \( x \) must be non-negative and small enough such that \( 36 - 2x \) is non-negative (since it represents the new length and width of the base of the box after the squares are cut out):
\[
36 - 2x \geq 0 \implies x \leq 18
\]
Also, \( x \) must be non-negative:
\[
x \geq 0
\]
Thus, the domain of \( V(x) \) is:
\[
0 \leq x \leq 18
\]

### Part (c): Using a table to estimate the range of the dimensions within which the volume is a maximum.

We can use the function \( V(x) \) to estimate the volume at different values of \( x \) within the domain. I will generate a table for values of \( x \) from 0 to 18 and calculate \( V(x) \).

### Part (d): Find the dimensions for the max volume and the max volume.

To find the dimensions that give the maximum volume, we need to find the maximum value of \( V(x) \) within the domain. This involves either analyzing the table from part (c) or taking the derivative of \( V(x) \), setting it to zero, and solving for \( x \).

Let's proceed by calculating the volume for several values of \( x \) to create a table for part (c) and then find the critical points for part (d) to determine the maximum volume and its corresponding dimensions.Here's the table showing the volume \( V(x) \) for \( x \) values from 0 to 18 cm:

\[
\begin{array}{|c|c|}
\hline
x \text{ (cm)} & V(x) \text{ (cm}^3\text{)} \\
\hline
0 & 0 \\
1 & 1156 \\
2 & 2048 \\
3 & 2700 \\
4 & 3136 \\
5 & 3380 \\
6 & 3456 \\
7 & 3388 \\
8 & 3200 \\
9 & 2916 \\
10 & 2560 \\
11 & 2156 \\
12 & 1728 \\
13 & 1300 \\
14 & 896 \\
15 & 540 \\
16 & 256 \\
17 & 68 \\
18 & 0 \\
\hline
\end{array}
\]

From the table, we observe that the volume reaches a maximum when \( x \) is around 6 cm, giving a volume of 3456 \( \text{cm}^3 \).

### Part (d): Finding Dimensions for Max Volume

To find the dimensions that provide the maximum volume, we use the maximum \( x \) value from the table, which is 6 cm. The base of the box will be:
\[
(36 - 2 \times 6) \text{ cm} \times (36 - 2 \times 6) \text{ cm} = 24 \text{ cm} \times 24 \text{ cm}
\]
Thus, the box with the maximum volume has dimensions 24 cm by 24 cm by 6 cm.

Do you need further details or have any questions about this solution? Here are some follow-up inquiries you might consider:

1. How does changing \( x \) affect the volume more precisely according to calculus?
2. What would happen if the size of the initial material was different?
3. Could we find the exact maximum volume using derivatives?
4. How does the thickness of the material affect the final volume?
5. What are practical considerations in making such a box in terms of material strength and foldability?

**Tip:** Always check the physical constraints and practicality of dimensions when designing real-world objects.

An open box is to be made from a square piece of material 36 cm on a side by cutting equal squares with a length of x from the corners and turning the sides up. a) Find the function V(x) for the volume. b) Determine the domain of V(x). c) Using a table to estimate the range of the dimensions within which the volume is a maximum. d) Find the dimensions for the max volume and the max volume.

Explore how to maximize the volume of an open box made from a square piece of material by cutting equal squares from the corners. This problem covers key concepts in algebra, geometry, and optimization, including volume calculation and domain determination. Suitable for students in Grades 10-12.

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