Let's break down and solve the questions given in this problem:

Problem Analysis

You start with a 12 cm by 12 cm square piece of cardboard. Square corners of side length $x$ cm are cut out from each corner, so that the sides can be folded up to form an open-top box. This configuration gives the box dimensions that depend on $x$, and we need to express the volume and answer the given questions based on it.

Solutions

1. Express the volume of the box as a function of $x$:

   After cutting out squares of side $x$ from each corner, the remaining base dimensions of the box are: $(12 - 2x) \text{ cm by } (12 - 2x) \text{ cm}$ The height of the box will be $x$ cm (the height after folding up the sides).

   The volume $V$ of a box is given by: $V = \text{length} \times \text{width} \times \text{height}$

   Substituting the dimensions: $V(x) = (12 - 2x)(12 - 2x)(x)$

   Expanding this expression: $V(x) = x(12 - 2x)^2$ $V(x) = x(144 - 48x + 4x^2)$ $V(x) = 4x^3 - 48x^2 + 144x$ So, the volume function is: $V(x) = 4x^3 - 48x^2 + 144x$

2. Find the domain of the function:

   Since the side length $x$ cannot be negative and must fit within the 12 cm by 12 cm square, the maximum possible value for $x$ occurs when $12 - 2x > 0$.

   Solving for $x$: $12 - 2x > 0$ $x < 6$ Additionally, $x$ must be greater than 0. Therefore, the domain of $x$ is: $0 < x < 6$

3. Determine the dimensions that yield the maximum volume using the graph:

   From the graph, we see that the volume is maximized when $x = 2$ cm, which gives a maximum volume of $V = 128$ cubic centimeters.

   Substituting $x = 2$ into the dimensions:

   • Length and width of the base: $12 - 2(2) = 12 - 4 = 8$ cm
   • Height: $x = 2$ cm

   Therefore, the dimensions that yield the maximum volume are: $\text{8 cm by 8 cm by 2 cm}$

Summary of Answers

• (a) The volume function: $V(x) = 4x^3 - 48x^2 + 144x$
• (b) Domain of $V(x)$: $0 < x < 6$
• (c) Maximum volume of 128 cm³ occurs at dimensions $8 \times 8 \times 2$ cm.

Would you like further details on any of these steps, or do you have any questions?

Here are some additional questions that could extend your understanding:

1. How would the volume formula change if the cardboard size was different (e.g., 15 cm by 15 cm)?
2. What would happen to the maximum volume if the cardboard had thickness?
3. Can you find the maximum volume using calculus (by differentiating the volume function)?
4. How would the domain change if we used rectangular cardboard instead of square?
5. How does changing $x$ impact each dimension of the box?

Tip: When optimizing dimensions, setting up a function and determining its derivative can help find maximum or minimum values quickly.