Problem 114 - College Algebra

Question

A machine produces open boxes using square sheets of metal. The machine cuts equal-sized squares measuring 3 inches on a side from the corners and then shapes the metal into an open box by turning up the sides. If each box must have a volume of 75 cubic inches, find the length of the side of the open square-bottom box.

Step-by-Step Solution

Answer

The length of each side of the square metal sheet is 11 inches.

Step 1: Represent the Problem With Variables and Equations

Since it's an open box, the length and width are the same for the base of the box, and the height is the side that's folded up. Let's take the original length of the square sheet as \( x \) inches. After cutting 3-inch squares from each corner, the new length and width will be \( (x - 6) \) inches each, and the height will equal 3 inches. The volume of an open box is given as \( length * width * height \). So in this case, the volume \( V \) is given as \( V = (x - 6) * (x - 6) * 3 \). We know that this volume is 75 cubic inches.

Step 2: Set up and Solve the Equation

Set up the equation: \( (x - 6) * (x - 6) * 3 = 75 \). Simplify it to \( x^2 - 12x + 36 = 25 \). Bring all the terms to one side to form a quadratic equation: \( x^2 - 12x + 36 - 25 = 0 \), which simplifies to \( x^2 - 12x + 11 = 0 \). Solve this quadratic equation by factoring or using the quadratic formula. The quadratic formula is \( x = [-b ± sqrt(b^2 - 4ac)] / (2a) \). In this equation, \( a = 1 \), \( b = -12 \), and \( c = 11 \).

Step 3: Find the Solutions and Identify the Length of the Side

Put these values into the quadratic formula to find the solutions. The solutions are \( x = 1 \) or \( x = 11 \). We disregard the solution \( x = 1 \) since the original metal sheet can't be 1 inch (we need to cut 3 inches from each side). Therefore, the length of each side of the square metal sheet is 11 inches.

Key Concepts

Algebra Word Problems

Algebra word problems are real-world problems that can be translated into mathematical equations using algebraic expressions and variables. Solving these problems often involves multiple steps: understanding the problem, identifying the relevant information, representing that information with variables, setting up equations, and finding the solution.

To tackle an algebra word problem, begin by carefully reading the problem to comprehend the scenario. In our example, square sheets of metal are used to make open boxes, with equal-sized squares removed from each corner. Knowing the volume that the box needs to achieve is also crucial. These details guide us in defining the variables and creating the equations. Let's denote the original length of the side of the square metal sheet as variable 'x'.

Once the information is represented in terms of variables, the next step is to translate the described actions (cutting out corners and folding the material) into an algebraic equation reflecting the final volume of the box. In our problem, after removing corners and folding, the new dimensions of the box become expressions in terms of 'x'. Equating the algebraic representation of the volume with the known volume allows us to find 'x' and, subsequently, the length of the side of the square-bottom box.

Volume of an Open Box

The volume of an open box can be visualized as the amount of space enclosed within it. For a box with a square base, the volume calculation involves three dimensions: the length of a side on the square base, the width (which is the same as the length for a square), and the height.

The formula for the volume of a rectangular prism, which applies to an open box, is length × width × height. When the sides are folded up after cutting out the corners, these dimensions change. In our problem, 3-inch squares are cut from each corner of the square metal sheets. So, if 'x' represents the original length of the side of the metal sheet, the new length and width will be 'x - 6' inches each (accounting for the removal of 3 inches from both sides of a corner), and the height is 3 inches.

Therefore, the equation to represent the volume will be \( V = (x - 6) \times (x - 6) \times 3 \). The goal is to find the value of 'x' that makes this volume exactly 75 cubic inches, which corresponds to the size of the box required.

Quadratic Formula

The quadratic formula is an essential tool in algebra for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The standard quadratic formula is \( x = [-b ± sqrt(b^2 - 4ac)] / (2a)\), where 'a', 'b', and 'c' are coefficients in the equation and 'sqrt' represents the square root.

In the context of our exercise, after setting up the volume equation, we simplify it and rearrange it into a quadratic equation: \( x^2 - 12x + 36 = 25 \), which then simplifies further to \( x^2 - 12x + 11 = 0\). Applying the quadratic formula, we identify 'a' as 1, 'b' as -12, and 'c' as 11. Plugging these into the formula, we find that 'x' can be either 1 or 11.

However, since we can't have the side of the original metal sheet to be 1 inch (as it wouldn't allow enough material to form the base after cutting out the corners), we can conclude that the correct solution is \( x = 11\) inches. This illustrates how the quadratic formula provides a systematic method to solve quadratic equations, even when factoring isn't straightforward.