To tackle this problem step-by-step, let's go through each part of the exercise based on the information provided:

Problem Analysis

1. We start with a square piece of cardboard with a width of 3 feet.
2. A square with side length $x$ is cut from each corner, and the sides are bent up to form an open-top box.

Given this setup, let:

• $x$: Side length of the squares being cut out.
• $y$: Length of the base of the box after folding.

Now, let's solve each part in turn.

Solution Steps

Part (a): Diagram and Volume Estimation

You would draw various configurations of the box by changing $x$, the side of the square being cut out. For each $x$, calculate the volume $V = y \times y \times x$, where $y$ is the length of the base after cutting out the corners.

Part (b): General Diagram

Draw a square with a smaller square (of side $x$) cut out at each corner. After folding, the height of the box becomes $x$, and the base length $y$ of the box becomes $3 - 2x$, as cutting $x$ from both sides reduces the width by $2x$.

Part (c): Expression for Volume in Terms of $x$ and $y$

Since the volume $V$ of a box is given by: $V = \text{base area} \times \text{height}$ we have: $V = y \times y \times x = y^2 \cdot x$ Thus, the volume $V$ can be expressed as: $V = x \cdot y^2$

Part (d): Relationship Between $x$ and $y$

Since $y$ represents the length of the base after cutting, and we start with a 3-foot wide piece of cardboard, we have: $y = 3 - 2x$ This equation expresses $y$ in terms of $x$.

Part (e): Volume as a Function of $x$ Only

Substitute $y = 3 - 2x$ into the volume equation: $V(x) = x \cdot (3 - 2x)^2$ Expanding $(3 - 2x)^2$ gives: $V(x) = x \cdot (9 - 12x + 4x^2)$ $V(x) = 4x^3 - 12x^2 + 9x$ So, the volume as a function of $x$ is: $V(x) = 4x^3 - 12x^2 + 9x$

Part (f): Maximizing the Volume

To find the maximum volume, take the derivative $V'(x)$ and set it to zero:

1. Differentiate $V(x) = 4x^3 - 12x^2 + 9x$.
2. Solve $V'(x) = 0$ for $x$ to find critical points.
3. Substitute back to find the maximum volume.

Would you like a detailed breakdown of finding the maximum volume in Part (f) using calculus?

Follow-Up Questions

1. How does changing the value of $x$ affect the height and base area of the box?
2. What are the physical limitations on the value of $x$ for this box construction?
3. How do we verify that a critical point provides a maximum volume and not a minimum?
4. What would happen to the maximum volume if the cardboard’s width changed from 3 feet to a different size?
5. Can you think of a real-world application for maximizing the volume of an open-top box in design or packaging?

Tip

When solving optimization problems, always check the domain of possible values for variables to avoid unrealistic results.