Problem 79 - College Algebra Essentials

Question

A rectangular lot whose perimeter is 360 feet is fenced along three sides. An expensive fencing along the lot's length costs \(\$ 20\) per foot and an inexpensive fencing along the two side widths costs only \(\$ 8\) per foot. The total cost of the fencing along the three sides comes to \(\$ 3280 .\) What are the lot's dimensions?

Step-by-Step Solution

Answer

The dimensions of the lot are Length = 33.34 feet and Width = 163.33 feet.

Step 1: Define variables

Let's denote the length of the rectangular lot as \(x\) feet, and the width as \(y\) feet.

Step 2: Formulate Equations

Using the information given in the problem, two equations can be formulated: 1. Perimeter equation: \( x + 2y = 360\) feet. This uses the information that three sides - one length (x) and two widths (2y) - are fenced and their total length is the given perimeter (360ft).2. Cost equation: \( 20x + 8*2y = 3280\) dollars. This equation uses the information about the cost per foot of the expensive and inexpensive fencing, and the total cost of the fencing.

Step 3: Rearranging the Perimeter Equation

To simplify the process, rearrange equation 1 to get \(x\) in terms of \(y\): \(x = 360 - 2y\).

Step 4: Substituting \(x\) in the Cost Equation

Substitute equation 3, into equation 2 which gives us: 20(360 - 2y) + 16y = 3280. This can be simplifed to: 7200 - 40y + 16y = 3280, which further simplifed gives: -24y = -3920.

Step 5: Solve for \(y\)

Divide both sides by -24, to get: \(y = 163.33\) feet.

Step 6: Solve for \(x\)

Substitute \(y = 163.33\) in equation 3 to get \(x\): \(x = 360 - 2*163.33 = 33.34\) feet.

Key Concepts

Algebraic Problem-Solving

Algebraic problem-solving is like a detective's investigation, where you look for clues, make deductions, and ultimately solve a mystery. In the case of our rectangular lot problem, the 'mystery' is finding the lot's dimensions.

The process involves defining variables for unknowns, in this case, the length ((x)) and width ((y)) of the lot. It involves gathering information from the problem statement and forming equations that reflect this information – the perimeter and cost.

Then, by manipulating these equations, such as rearranging terms and substituting values from one equation into another, you can uncover the values of the variables. This can require several steps, including simplifying expressions, isolating terms, and executing operations such as addition, subtraction, multiplication, or division.

A key skill in algebraic problem-solving is the ability to translate a word problem into mathematical language and equations. This often involves identifying and discarding irrelevant information, focusing only on what is necessary to solve the problem at hand.

Perimeter Calculation

Calculating the perimeter of a shape involves adding up the lengths of all its sides. For rectangles, the formula is straightforward: you double the sum of the length and width (2(l + w)). However, our problem presents a twist: the perimeter accounts for only three sides.

To adjust, we use the formula (l + 2w), representing one length and two widths. This is because our problem states that only three sides of the rectangular lot are fenced.

Understanding perimeter calculation is essential because it translates the physical constraints of the problem into an algebraic equation. In this case, it serves as our first equation, helping us realize the relationship between the length and width.

It is crucial to visualize the problem when dealing with geometrical shapes. Visualizing not only helps in understanding the problem but also avoids common mistakes, like calculating the perimeter of four sides when only three sides are given.

Linear Equations in Algebra

Linear equations are the backbone of algebra and represent relationships where each variable is to the power of one. These equations appear as straight lines when plotted on a graph and often have the standard form (y = mx + b), where (m) is the slope, and (b) is the y-intercept.

In our problem, after setting up our perimeter and cost equations, we see a system of linear equations, one of which we rearranged to express (x) in terms of (y). This is a technique called substitution, used to solve for one variable first, making it easier to tackle the problem.

The strength of using linear equations lies in their simplicity and the practical tools they offer for solving real-world problems. By understanding how to manipulate these equations, students unlock the ability to solve a vast array of problems across different disciplines.

Remember, when dealing with equations, each transformation you make should keep the equation balanced. It's like a scale: what you do to one side, you must also do to the other. This balance is key to maintaining the equation's integrity and arriving at the correct solution.