Problem 54 - Algebra for College Students

Question

Set up an equation and solve each problem. A rectangular piece of cardboard is 2 units longer than it is wide. From each of its corners a square piece 2 units on a side is cut out. The flaps are then turned up to form an open box that has a volume of 70 cubic units. Find the length and width of the original piece of cardboard.

Step-by-Step Solution

Answer

The width is 9 units, and the length is 11 units.

Step 1: Define Variables

Let's define the variables. Let the width of the original piece of cardboard be \( w \) units. Because the cardboard is 2 units longer than it is wide, the length will be \( w + 2 \) units.

Step 2: Define the Dimensions after Cutting

Since squares with side 2 are cut from each corner, the new dimensions of the base of the box are reduced by 4 units in both the length and width directions. Thus, the new width is \( w - 4 \) and the new length is \( (w + 2) - 4 = w - 2 \).

Step 3: Define Volume Equation

The volume of the box is given by the equation: \[ V = \text{Length} \times \text{Width} \times \text{Height} \]Given the box height is 2 (the dimension of the squares cut out), the length is \( w - 2 \), and the width is \( w - 4 \), the volume equation becomes:\[ 70 = (w - 2) \cdot (w - 4) \cdot 2 \]

Step 4: Simplify and Solve the Equation

First, simplify the volume equation: \[ 70 = 2(w - 2)(w - 4) \]Divide both sides by 2 to simplify:\[ 35 = (w - 2)(w - 4) \]Now expand the equation:\[ 35 = w^2 - 4w - 2w + 8 \]Simplify to get:\[ 35 = w^2 - 6w + 8 \]Subtract 35 from both sides:\[ 0 = w^2 - 6w - 27 \]Solve the quadratic equation using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -6, c = -27 \).

Step 5: Calculate Using the Quadratic Formula

Substitute the coefficients into the quadratic formula:\[ w = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1} \]Simplify within the formula:\[ w = \frac{6 \pm \sqrt{36 + 108}}{2} \]\[ w = \frac{6 \pm \sqrt{144}}{2} \]\[ w = \frac{6 \pm 12}{2} \]This gives us two potential solutions: \( w = 9 \) or \( w = -3 \). Since width cannot be negative, \( w = 9 \).

Step 6: Find Length Using Calculated Width

Substitute \( w = 9 \) into the expression for the length:\( \text{Length} = w + 2 = 9 + 2 = 11 \).

Step 7: Verify the Solution

Verify the measurements form a box with correct volume:After cutting squares of 2 units, the resulting dimensions are width \( 5 \), length \( 7 \), height \( 2 \).Calculate volume: \( 5 \times 7 \times 2 = 70 \) cubic units.This matches the given volume, confirming the solution is correct.

Key Concepts

Quadratic Equations

In algebra, a quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). Solving quadratic equations is a fundamental skill. Here, the quadratic was used to solve for the width of a cardboard.
Quadratic equations can be solved through various methods:

• Factoring: Used when the quadratic is easily factored, making it possible to find solutions by setting each factor equal to zero.
• Completing the Square: Involves rewriting the quadratic in the form \((x+p)^2=q\), which can then be solved for \(x\).
• Quadratic Formula: The most universally applicable method; it works even when other methods don't. The formula is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In the solution, the quadratic formula was applied because the equation \( w^2 - 6w - 27 = 0 \) did not factor easily. After calculating with the formula, \( w = 9 \) was determined as the width, since the width cannot be negative.

Volume Calculation

Volume is the measure of the space occupied by a 3-dimensional object, expressed in cubic units. For a box (rectangular prism), the volume is calculated using the formula: \( V = \text{Length} \times \text{Width} \times \text{Height} \).

In this exercise, an open-top box formed by cutting squares from the corners of a rectangular cardboard and folding its sides up was considered. After cutting and folding, the dimensions of the box were known:

• Height: Matching the side of the square cut out, which was 2 units.
• Width: Calculated as \( w - 4 \).
• Length: Calculated as \( w - 2 \).

With these expressions, substituting into the volume formula allowed setting up the equation \( 70 = (w - 2)(w - 4)\cdot2 \). Dividing through by 2 helped simplify the calculations before solving for \( w \), highlighting how knowing the volume guides solving real-world problems by back-tracking through algebra.

Rectangular Dimensions

Understanding dimensional transformations of rectangular shapes is key to solving problems involving cutting and rearranging shapes, as seen in this exercise.

Initially, we had dimensions \( w \) (width) and \( w + 2 \) (length) for the cardboard. Upon removing squares from each corner:

• Length became \( w - 2 \).
• Width became \( w - 4 \).

These transformations are critical for setting up the volume equation because they represent how the cardboard transitions from flat to a constructed 3D shape.
When given the task of altering dimensions in any problem, always:

• Clearly define the original size.
• Account for reductions or changes due to cutting or folding.
• Check that dimensions logically follow from the operations performed.

This ensures the accurate formation of both the equation and subsequent solution derivation. The original width was found to be 9 units, making the length 11 units, computed by simply adding 2 units to the width.