Problem 23 - Precalculus Student Solutions Manual 5th

Question

A rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four corners, and the flaps are folded upward to form an open box. If the volume of the box is 832 in. \(^{3}\), what were the original dimensions of the piece of metal?

Step-by-Step Solution

Answer

Width: 20 in, Length: 30 in

Step 1 - Define Variables

Let the width of the original piece of metal be represented by \(w\) inches. Thus, the length of the piece of metal is \(w + 10\) inches, because the length is 10 inches longer than the width.

Step 2 - Consider the Dimensions After Cutting Squares

After cutting the 2-inch squares from each corner, the new dimensions of the base of the box will be \(w - 4\) inches for the width and \(w + 10 - 4 = w + 6\) inches for the length. The height of the box will be the height of the squares that were cut out, which is 2 inches.

Step 3 - Write the Volume Formula

The volume of the box can be described using the formula for the volume of a rectangular prism: \(V = l \times w \times h\). Substituting the dimensions, the equation becomes: \[ \text{Volume} = (w - 4)(w + 6)(2) \]

Step 4 - Substitute Given Volume

We know the volume of the box is 832 cubic inches. Substituting this value into the volume formula gives: \[ 2(w - 4)(w + 6) = 832 \]

Step 5 - Simplify the Equation

Divide both sides by 2 to simplify: \[ (w - 4)(w + 6) = 416 \]

Step 6 - Expand and Solve Quadratic Equation

Expand the left side of the equation to form a quadratic equation: \[ w^2 + 2w - 24 = 416 \] Subtract 416 from both sides: \[ w^2 + 2w - 440 = 0 \]

Step 7 - Solve the Quadratic Equation

Use the quadratic formula \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1, b = 2, c = -440\). Substitute these values: \[ w = \frac{-2 \pm \sqrt{2^2 - 4(1)(-440)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 1760}}{2} = \frac{-2 \pm \sqrt{1764}}{2} \] Simplify further: \[ w = \frac{-2 \pm 42}{2} \] Therefore, the possible solutions for \(w\) are: \[ w = 20 \] or \[ w = -22 \] Since the width cannot be negative, \(w = 20\).

Step 8 - Find Original Dimensions

Given \(w = 20\) inches, the original width is 20 inches and the original length is \(w + 10 = 20 + 10 = 30\) inches.

Key Concepts

quadratic equation

In our problem, we use a quadratic equation to find the width of the original piece of metal. A quadratic equation is a second-degree polynomial. It generally has the form \[ax^2 + bx + c = 0\]. In our case, to find the width \(w\), we had to solve the equation: \[w^2 + 2w - 440 = 0\]. To solve this, we used the quadratic formula: \[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. By substituting the values \(a = 1\), \(b = 2\), and \(c = -440\), we found that the possible widths are 20 inches or -22 inches. Since a width cannot be negative, we concluded that \(w = 20\).

rectangular dimensions

The dimensions of a rectangle are described by its length and width. In this problem, we start with a piece of metal and define its width as \(w\) inches and its length as \(w + 10\) inches. We assume its length is 10 inches longer than its width. After cutting out squares from the corners, the dimensions of the remaining part that forms the base of the box are \(w - 4\) inches (width) and \(w + 10 - 4 = w + 6\) inches (length). These new dimensions account for the loss of 4 inches (2 inches from each corner). Understanding how dimensions change is crucial for solving complex geometry problems.

volume calculation

To solve the problem, we calculate the volume of the open box formed after cutting and folding the metal piece. The volume \(V\) of a rectangular prism is given by the formula: \[V = l \times w \times h\], where \(l\) is the length, \(w\) is the width, and \(h\) is the height. For this box, the height is 2 inches (height of the cut squares). By substituting the dimensions and the given volume of 832 cubic inches, we form the equation: \[2(w - 4)(w + 6) = 832\]. We then simplify and solve this equation to find the width of the original piece.

problem solving

Solving this problem involves multiple steps. It illustrates a structured approach to problem-solving:

• Define variables clearly: let \(w\) be the width.
• Account for dimension changes: new width \(w - 4\) and length \(w + 6\).
• Set up relevant formula: volume of the box \(V = 2(w - 4)(w + 6) = 832\).
• Simplify and solve the equation: derive and solve a quadratic equation.
• Apply logical reasoning: Identify physically reasonable (positive) solutions.

By following these structured steps, we ensure a comprehensive and accurate solution to the problem.