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: