Problem 54 - Intermediate Algebra

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 original dimensions of the cardboard are 9 units in width and 11 units in length.

Step 1: Define Variables

Let the width of the rectangular piece of cardboard be \( w \) units. Since the length is 2 units longer than the width, we can express the length as \( w + 2 \) units.

Step 2: Understand the Box Formation

From each corner, a square of side 2 units is cut. The height of the box formed will thus be 2 units, as that is the side length of the square cut out.

Step 3: Set Up Volume Equation

The volume of the box, when the squares are cut and the flaps are turned up, is given as 70 cubic units. The new dimensions of the base of the box are \((w - 4)\) units in width and \((w - 2)\) units in length, after accounting for the cut-out squares. Therefore, the volume equation is:\[ (w - 4)(w - 2)(2) = 70 \]

Step 4: Simplify the Equation

Simplify the volume equation by first dividing both sides by 2:\[ (w - 4)(w - 2) = 35 \]

Step 5: Solve the Quadratic Equation

Expand and simplify:\[ w^2 - 6w + 8 = 35 \]Re-arrange to form a standard quadratic equation:\[ w^2 - 6w - 27 = 0 \]Solve using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = -27 \). This gives:\[ w = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2} \]

Step 6: Evaluate Solutions

The potential solutions are \( w = 9 \) and \( w = -3 \). Since a width cannot be negative, the width \( w = 9 \) units is the only viable solution.

Step 7: Determine Original Dimensions

If \( w = 9 \), then the original width is 9 units, and the length is \( w + 2 = 11 \) units.

Key Concepts

Volume of a Box

The volume of a box is a fundamental concept in geometry. It represents the amount of space inside the box. In simpler terms, think of it as how much you can fill a box with something, say sand or water.

For rectangular boxes, which are simply called rectangular prisms, the volume is calculated by multiplying its length, width, and height. These three dimensions should be in the same units. Here's the formula for a box’s volume:

• Volume (\( V \) ) = \( ext{Length} imes ext{Width} imes ext{Height} \)

In our exercise, we formed a box by cutting squares from each corner of a cardboard and folding the flaps upwards. This affects the dimensions of the base of the box. The new width and length become:

• Adjusted Width = Original Width - 4 units.
• Adjusted Length = Original Length - 2 units.

With the given volume of 70 cubic units, setting up an equation helps us find the original dimensions. This process lays the groundwork for solving many practical box-related problems.

Quadratic Equations

Quadratic equations are vital in mathematics and appear when dealing with problems involving squared terms. A standard quadratic equation looks like this:
\( ax^2 + bx + c = 0 \).

• \( a, b, \) and \( c \) are constants, with \( a eq 0 \).

In our case, we dealt with a quadratic equation while finding the width of the original cardboard piece. After simplifying the dimensions into a standard form, we got:
\( w^2 - 6w - 27 = 0 \).Solving the Quadratic Equation
You can solve quadratic equations using various methods, such as factoring, completing the square, or the quadratic formula. Here, we used the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).This formula is handy when factoring is challenging or impossible. In the exercise, the quadratic formula helped us find potential widths for the cardboard, ultimately identifying the feasible solution that fits the problem context.

Geometric Problem Solving

Geometric problem solving involves applying geometric concepts to figure out unknown dimensions or properties within a shape. It often combines known formulas and creative thinking.

For the cardboard box exercise, we used geometric problem-solving skills as follows:

• Identifying changes in dimensions due to cutting corners.
• Setting up a relationship between the new dimensions and the box's volume.
• Connecting geometric adjustments to algebraic equations.

Practical Approach
We applied problem-solving strategies to redefine dimensions after cutting squares from each corner. This illustrates a common task in geometric exercises: modifying a shape and determining how those modifications influence the overall properties. Critical to this is understanding how each action (like cutting the corners) impacts the overall shape, which then must be translated into mathematical expressions to solve the problem efficiently. This methodological approach to problem solving is essential in tackling real-life geometric challenges.