To solve this problem, let's analyze the given information step-by-step.

1. Initial Cardboard Dimensions:

   • The length of the cardboard is three times the width.
   • Let's denote the width of the cardboard as $w$.
   • Therefore, the length of the cardboard is $3w$.

2. Cutting the Corners:

   • Squares of side 15 cm are cut from each corner of the cardboard.

3. Dimensions after Cutting:

   • The new dimensions of the cardboard after cutting and folding:
     • New width = $w - 2 \times 15$ = $w - 30$
     • New length = $3w - 2 \times 15$ = $3w - 30$
     • Height of the box = 15 cm (since the corners are folded up to form the sides of the box).

4. Volume of the Box:

   • Volume $V$ of the box is given by: $V = \text{length} \times \text{width} \times \text{height}$ Substituting the dimensions: $V = (3w - 30)(w - 30) \times 15$

5. Simplifying the Volume Function:

   • Expand the expression: $V = 15 (3w - 30)(w - 30)$ $V = 15 (3w^2 - 90w - 30w + 900)$ $V = 15 (3w^2 - 120w + 900)$ $V = 45w^2 - 1800w + 13500$

Thus, the volume $V$ of the open box is given by: $V = 45w^2 - 1800w + 13500$

6. Nature of the Resultant Function:
   • The function $V = 45w^2 - 1800w + 13500$ is a quadratic polynomial function in terms of the variable $w$.

Would you like any further details or have any questions?

Here are some related questions you might be interested in:

1. How do you determine the dimensions of a box given different cutting lengths?
2. What are the characteristics of polynomial functions?
3. How can we maximize the volume of the box by choosing the appropriate dimensions?
4. What is the impact of changing the height on the volume of the box?
5. How does the shape of the original cardboard affect the final box dimensions?

Tip: Always verify the dimensions after each modification step to ensure correctness in multi-step problems like this one.