Question

Use the following information. You own a bottle recycling center that receives bottles that are either sorted by color or unsorted. To sort and recycle all of the bottles, you can use up to 4200 hours of human labor and up to 2400 hours of machine time. The system below represents the number of hours your center spends sorting and recycling bottles where \(x\) is the number of tons of unsorted bottles and \(y\) is the number of tons of sorted bottles. \(4 x+y \leq 4200\) \(2 x+y \leq 2400\) \(x \geq 0, y \geq 0\) a. Find the vertices of your graph. b. You will earn 30 dollar per ton for bottles that are sorted by color and earn 10 dollar per ton for unsorted bottles. Let \(P\) be the maximum profit. Substitute the ordered pairs from part (a) into the following equation and solve. $$ P=10 x+30 y $$ c. Assuming that the maximum profit occurs at one of the four vertices, what is the maximum profit?

Short answer

The maximum profit can be found after all the vertices are calculated and each is substituted into the profit function \(P = 10x + 30y\). The solution will be the vertex that results in the highest value for P.

Step-by-step solution

- Identify the Constraints and Plot Them

The equations given are \(4x + y \leq 4200\) and \(2x + y \leq 2400\). We first graph these equations, treating them as equalities. The graph will be a plot on (x, y) plane with the area bounded by these lines and the x and y axes falling within the feasible region.

- Finding the vertices

Now, we find intersection points of the lines, as well as their intersections with the axes. These points are termed vertices. To find the intersection of two lines, equate them and solve for x and y simultaneously. The intersection points for axes can be found by setting the variables to 0 in each of the equations. Solve the equations to find the coordinates of the vertices.

- Substituting the coordinates in the Profit equation

The profit function is given by the equation \(P = 10x + 30y\). Once the vertices are found, substitute each pair in the place of x and y into the profit equation in order to compute and compare the different possible profits.

- Identify maximum profit

The maximum profit will be the greatest value obtained from the profit equation in Step 3. Thus you will see which point (and therefore which values of x and y- tons of unsorted and sorted bottles respectively) gives you the maximum profit out of all. This point is the solution of our linear programming problem.

Key concepts

System of Inequalities

Imagine you're running a recycling center and the key to maximizing profits is efficient resource allocation, governed by a system of inequalities. These inequalities are like a recipe with specific limits on ingredients—in this case, the hours of human labor and machine time available for processing bottles.

In the context of algebra, a system of inequalities represents multiple inequalities that are all true at the same time. For the recycling center, we have two crucial inequalities: \(4x + y \leq 4200\) and \(2x + y \leq 2400\), where \(x\) and \(y\) correspond to the tons of unsorted and sorted bottles, respectively. These inequalities form a set of conditions that must be met simultaneously to keep operations viable.

A well-understood system of inequalities enables you to visualize the feasible region—a space where all the constraints are satisfied. Here, it's representing possible combinations of hours spent sorting different types of bottles while staying within labor and machine hour limits. Understanding this system is crucial for making data-driven decisions that align with business capabilities.

Graphing Linear Inequalities

Graphing linear inequalities is like drawing a map that guides you to the treasure—maximum profit. The first step to interpret the system of inequalities is transforming them into a visual plane. Each inequality is graphed as if it were an equation. To graph \(4x + y = 4200\) and \(2x + y = 2400\), you'd plot straight lines representing these equations on a coordinate system.

Here's where we need to consider the nature of inequalities. Unlike equations, inequalities do not describe just a line, but everything on one side of that line as well. The entire area that lies below or above this line corresponds to the set of all solutions that satisfy the inequality. After graphing, you will shade the region that contains solutions to both inequalities, which is known as the feasible region.

Understanding the graph is paramount as it visualizes not only the restrictions placed by the inequalities but also displays potential areas for optimization, guiding you to make the most educated and efficient decisions for operations.

Optimization Problem

At the heart of every business challenge lies an optimization problem, aiming to make the best out of available resources. In terms of mathematics and our recycling center scenario, the goal is achieving maximum profit, given the constraints.

These kinds of problems typically ask, 'What is the most profitable way to use my limited time and resources?'. In the case of the inequalities describing labor and machine hours, we are seeking to maximize profits while staying within these bounds.

Applying algebraic methods, we use the inequalities to sketch a feasible region and then employ the profit function \(P = 10x + 30y\) to evaluate each corner point of that region. The corner points, or vertices, are where we often find our maximum or minimum values because they represent the extremities of our feasible region. By calculating the profit for each vertex, we can determine where our operations will be most fruitful. This process turns a complex business scenario into a manageable mathematical model.

Maximum Profit Calculation

Now, let's crunch the numbers to calculate the maximum profit. It's like finding the highest peak in a mountain range; we assess each vertex to identify where we hit the ceiling on what we can earn.

Once we have our system graphed and the feasible region determined, we take the coordinates of each vertex and plug them into the profit equation, \(P = 10x + 30y\). This action gives us the total potential profit for each combination of sorted and unsorted bottles.

By comparing the calculated profits for each vertex, the largest figure that emerges is our maximum potential profit—this is the ultimate goal of our optimization problem. Ensuring that the calculation is correct and that all potential vertices are checked is essential because missing even one could lead us to settle for a lower peak than we could have actually reached. Profit optimization in algebra isn't just a theoretical exercise, it's a fundamental business strategy that can yield tangible, impactful results.