Problem 26 - Algebra Form and Function

Question

American Airlines limits the size of carry-on baggage to 45 linear inches (length \(+\) width \(+\) height \()\), with a weight of no more than 40 pounds. \({ }^{1}\) (a) If the length and width of a piece of luggage both measure \(x\) inches, express the maximum height of the luggage in terms of \(x\). (b) Express the volume of the piece of luggage in part (a) in terms of \(x\). (c) Find the zeros of your equation from part (b). What does this tell you about the dimensions of the piece of luggage?

Step-by-Step Solution

Answer

Answer: The zeros of the volume equation are 0 and \(45/2\). They represent the points where the volume of the piece of luggage becomes zero, indicating that these dimensions are not possible for a piece of luggage with a non-zero volume.

Step 1: (a) Express the maximum height in terms of x

Let's first consider the dimensions of the luggage. Length (\(L\)), width (\(W\)), and height (\(H\)) are given with the relation \(L + W + H \leq 45\). Given that the length and width both measure \(x\) inches, the equation becomes \(x + x + H \leq 45\). So, we first find an expression for \(H\): \begin{align*} 2x + H &\leq 45 \\ H &\leq 45 - 2x \end{align*} Now we have an expression for the maximum height: \(H \leq 45 - 2x\).

Step 2: (b) Express the volume in terms of x

To find the volume of the luggage, we use the formula \(V=LWH\). Since the luggage has a length and width of \(x\) inches, and the maximum height is \(45-2x\), the volume can be calculated as: \[V = x \cdot x \cdot (45 - 2x)\] which simplifies to: \[V = x^2 (45 - 2x)\] Now we have the volume of the piece of luggage in terms of \(x\): \[V = x^2(45 - 2x)\].

Step 3: (c) Find the zeros of the volume equation

To find the zeros of the volume equation \(V = x^2(45 - 2x)\), we find the values of \(x\) for which the volume is 0. We set the expression equal to 0 and solve for x: \begin{align*} x^2(45 - 2x) &= 0 \end{align*} This equation has two factors, \(x^2\) and \((45 - 2x)\). The zeros occur when either factor is equal to 0. 1. \(x^2 = 0 \Rightarrow x = 0\) 2. \(45 - 2x = 0 \Rightarrow x = \frac{45}{2}\) The zeros of the volume equation are 0 and \(\frac{45}{2}\). This means that the volume becomes zero when \(x = 0\) or \(x = \frac{45}{2}\). These zeros tell us that if the length and width of the luggage are equal to 0 or \(\frac{45}{2}\), the volume will become zero, which means that these dimensions are not possible for a piece of luggage with a non-zero volume.

Key Concepts

Volume Calculation

To calculate the volume of a three-dimensional object like luggage, we use the formula for volume, which for a rectangular prism is given by:\[ V = L imes W imes H \]Here, \(L\) is the length, \(W\) is the width, and \(H\) is the height. In the problem, both length and width are expressed in terms of \(x\), resulting in a dimension of \(x \times x\). The maximum height of our luggage is represented as \(H = 45 - 2x\) because the total dimension (length + width + height) cannot exceed 45 inches. Thus, the volume equation becomes:\[ V = x^2 (45 - 2x) \]
The exercise helps emphasize the relationship between dimensions and volume, illustrating how they are intrinsically linked. If either increases, the overall volume can change dramatically, showcasing a fundamental principle in geometry.

Equation Zeros

In mathematics, the zeros of an equation are the values for which the function equals zero. For our luggage volume, the equation is:\[ V = x^2 (45 - 2x) \]To find the zeros, we set this equation equal to zero and solve for \(x\):

• The first factor, \(x^2 = 0\), implies \(x = 0\).
• The second factor, \(45 - 2x = 0\), implies \(x = \frac{45}{2}\).

These zeros tell us that at \(x = 0\), the dimensions collapse into a flat, two-dimensional plane, essentially eliminating any volume. At \(x = \frac{45}{2}\), the piece of luggage reaches its maximum possible width and length such that the remaining dimension (height) equals zero, resulting in a zero volume. Therefore, the zeros are crucial in understanding the possible limits and constraints of the physical dimensions.

Dimensions in Algebra

Dimensions in algebra related to real-world objects like luggage are key in creating meaningful expressions and constraints. The exercise exemplifies how algebra can model real situations, using dimensions to create expressions like:

• The linear equation \(L + W + H \leq 45\), setting limits.
• Algebraic manipulations such as \(H = 45 - 2x\) to derive height.
• The quadratic volume equation \(V = x^2 (45 - 2x)\), illustrating relationships.

Such algebraic expressions are instrumental in problem-solving, allowing us to translate constraints into mathematical language. This approach underscores the relevance of algebra in designing and understanding practical constraints, where each dimension must be considered to achieve the desired outcome, like ensuring the luggage doesn't exceed a certain size while maintaining enough volume for packing.