Problem 39 - A History of Mathematics: An Introduction

Question

Solve the following problem from tablet AO 8862: Length and width. I combined length and width and then I built an area. I turned around. I added half of the length and a third of the width to the middle of my area so that it was \(15 . \mathrm{I}\). returned. I summed the length and width and it was 7 . What are the length and width?

Step-by-Step Solution

Answer

Answer: The length (L) of the rectangle is 4 and the width (W) is 3.

Step 1: Define the variables and the equations.

Let's call the length of the rectangle L and the width W. We are given the following relationships: 1. The sum of the length and the width is equal to 7 (L + W = 7) 2. Half of the length and a third of the width, added to the area, is equal to 15 (L * W + (L/2) + (W/3) = 15) Now we have two equations with two variables, L and W: (1) L + W = 7 (2) L * W + (L/2) + (W/3) = 15

Step 2: Solve the first equation for one of the variables.

We will solve equation (1) for one of the variables, in this case, let's solve for L: L = 7 - W

Step 3: Substitute the expression for L in the second equation.

Now, we will substitute the expression for L that we found in step 2 into equation (2): (7 - W) * W + ((7 - W) / 2) + (W / 3) = 15

Step 4: Solve the equation for the remaining variable.

Now we have an equation with only one variable, W. Let's solve this equation: (7 - W) * W + ((7 - W) / 2) + (W / 3) = 15 Multiplying both sides of the equation by 6 to clear fractions: 6 * (7 - W) * W + 3 * (7 - W) + 2 * W = 90 Expanding the equation: 42W - 6W^2 + 21 - 3W + 2W = 90 Combining like terms: -6W^2 + 41W - 69 =0

Step 5: Solve the quadratic equation for W.

We have a quadratic equation in the form of ax^2 + bx + c = 0, which in this case is -6W^2 + 41W - 69 = 0. Let's solve for W: Using the quadratic formula, W = (-b ± sqrt(b^2 - 4ac)) / 2a: W = ( -41 ± sqrt( 41^2 - 4(-6)(-69) )) / 2(-6) W = (-41 ± sqrt(1681 - 1656)) / (-12) W = (-41 ± sqrt(25)) / (-12) The two possible solutions for W are: W = (-41 - 5) / (-12) = 46 / 12 = 23 / 6 W = (-41 + 5) / (-12) = 36 / 12 = 3

Step 6: Determine the reasonable solution for W and find L.

Since W must be less than 7 (as L + W = 7), the reasonable solution for W is 3. Now we can find L using the expression we found in step 2: L = 7 - W L = 7 - 3 L = 4

Step 7: State the solution.

The length (L) of the rectangle is 4 and the width (W) is 3.

Key Concepts

Algebraic Problem-Solving

Algebraic problem-solving is a method of finding unknown values by using mathematical operations and expressions. In the given textbook problem, we dealt with a classic algebra problem: finding the dimensions of a rectangle given certain conditions.

The first step was to define the variables (length and width) and establish equations based on the given relationships. This involved translation of worded problems into algebraic expressions, an essential skill in mathematical reasoning. In this case, the relationships described involved the perimeter and area modifications of the rectangle.

Once the equations were set up, we applied a systematic approach to isolate one variable and then substituted it into the other equation. This deft maneuver is part of the algebraic toolbox that allows us to whittle down complex problems into simpler ones we can solve.

Mathematical Reasoning

Mathematical reasoning involves the logical deduction and strategic planning required to solve mathematical problems. Our textbook example demonstrates this through several steps. For instance, we assessed which equation to solve first and how to manipulate the expressions in a way that would simplify the problem.

This evident strategic thinking is the hallmark of mathematical reasoning. By multiplying both sides of the equation by 6 (Step 4), fractions were eliminated, simplifying the calculations and leading to a quadratic equation. The selection of the right value for the width (W) post solving the quadratic equation also involved mathematical reasoning, as it required understanding the real-world implications of the solution, wherein a width larger than the sum of both dimensions is nonsensical.

Ancient Mathematics

The problem given from tablet AO 8862 is not just a mathematical brainteaser; it’s a snippet of ancient mathematics at work. The mathematics of early civilizations, like the Babylonians, often dealt with geometric and algebraic problems related to surveying, construction, and trade.

In this sense, the problem stems from a real-world scenario one might have encountered thousands of years ago. It's fascinating to realize that the algebraic methods we used for solving the problem—defining variables, setting up equations, and solving quadratics—are rooted in techniques that have been developed and refined over millennia. This serves as a reminder of the timelessness and universality of mathematical principles.