Problem 69 - College Algebra

Question

An open box is to be constructed by cutting out square corners of \(x\) -inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of \(x\)

Step-by-Step Solution

Answer

Volume function: \( V(x) = 4x^3 - 32x^2 + 64x \).

Step 1: Understanding the Problem

We need to find the volume of an open box. The box is formed by cutting out squares of side length \(x\) in a piece of cardboard that is initially 8 inches by 8 inches and folding up the sides.

Step 2: Defining Dimensions of Box

After cutting small square corners of \(x\), the length and width of the base of the box will both reduce by \(2x\). Thus, the new dimensions of the base will be \((8 - 2x) \times (8 - 2x)\). The height of the open box will be \(x\) since the depth is equal to the side of the squares.

Step 3: Expressing Volume as a Function

Volume \(V\) of a box is given by \(\text{length} \times \text{width} \times \text{height}\). For our box, this is \(V(x) = (8 - 2x)(8 - 2x)x\). To simplify, this becomes \( V(x) = x(8 - 2x)^2 \).

Step 4: Simplifying the Volume Function

We simplify \( V(x) = x(8 - 2x)^2 \). First, expand \((8 - 2x)^2\) resulting in \(64 - 32x + 4x^2\). Then multiply by \(x\), giving \( V(x) = x(64 - 32x + 4x^2) = 4x^3 - 32x^2 + 64x\).

Key Concepts

Polynomial Functions

The volume of our open box can be expressed as a polynomial function of the variable \(x\). In mathematics, polynomial functions are equations composed of variables raised to powers combined with constants through operations of addition, subtraction, and multiplication. A key feature of polynomial functions is that the variable exponents must be whole numbers.

When expressing the volume, we expanded and simplified \((8 - 2x)^2\) and then multiplied it by \(x\) to arrive at our polynomial \(V(x) = 4x^3 - 32x^2 + 64x\). Here, the highest exponent is 3, making this a cubic polynomial function. Polynomials are very useful for modeling because they can describe complex relationships in simple algebraic terms.

Because the terms of the polynomial are ordered by their degree, the leading term \(4x^3\) significantly influences the function's behavior as \(x\) increases or decreases. Thus, understanding polynomial characteristics can give us insights into how the box's volume changes as we adjust \(x\).

Geometric Dimensions

Understanding the geometric dimensions of our problem is crucial for correctly deriving the volume function. We started with an 8-inch by 8-inch piece of cardboard. By cutting out squares of side \(x\) from each corner, we modified the base dimensions and determined the box's height.

• The length and width of the base are each reduced by \(2x\), leading to new dimensions of \((8 - 2x)\) by \((8 - 2x)\).
• The height of the box is \(x\), which comes from the depth of the cuts made.

To find the volume of the open box, we use the formula for the volume of a rectangular box, \(V = \text{length} \times \text{width} \times \text{height}\). With this setup, it becomes \(V(x) = (8 - 2x)(8 - 2x)x\). Understanding these dimensions not only helps in setting up the problem correctly but also in visualizing how the box structure is formed and changed.

Function Notation

Function notation is a succinct way to represent mathematical relationships. It's particularly useful when dealing with equations that describe how one quantity depends on another, such as in our box volume problem.

In function notation, \(V(x)\) represents the volume of the box as a function of \(x\), where \(x\) is the size of the squares cut from each corner of the cardboard. This notation tells us two things:

• \(V\) is dependent on \(x\), signifying a relationship between the volume and the size of the cut squares.
• Any value substituted for \(x\) returns the corresponding volume for that specific square size.

This neat packaging of information allows us to efficiently manipulate and analyze relationships without repeatedly describing each variable's dependency. Utilizing function notation, such as \(V(x) = 4x^3 - 32x^2 + 64x\), aids in understanding and communicating complex relationships in a concise format.