Problem 66 - Beginning and Intermediate Algebra

Question

A machine in a factory can be calibrated to fill either large or small bags of potato chips. The machine will run at most 12 hr per day. Let \(x=\) number of hours the machine fills large bags \(y=\) number of hours the machine fills small bags a) Write a system of linear inequalities to describe the constraints on the number of hours the machine fills the bags each day. b) Graph the feasible region that describes how the hours can be distributed between filling the large and small bags of chips. c) Find three points in the feasible region and discuss their meanings. d) Find one point outside the feasible region and discuss its meaning.

Step-by-Step Solution

Answer

In summary, we have a system of linear inequalities representing the constraints on the number of hours the machine fills large and small bags, as follows: \(x+y\leq 12\), \(x\geq 0\) and \(y\geq 0\). The feasible region is a triangle with vertices at (0,0), (0,12), and (12,0). Three points within the feasible region, such as (2,6), (4,4), and (6,2), represent different ways of distributing working hours, while an example of a point outside the feasible region, (10,10), exceeds the maximum allowed working hours for the machine.

Step 1: Write the constraint that covers how long the machine can run

Let \(x\) represent the number of hours the machine fills large bags, and \(y\) represent the number of hours the machine fills small bags. The machine can run at most 12 hours per day, so we can write the constraint as: \[x + y \leq 12\] Also, keep in mind that the number of hours filling each type of bag must be non-negative. We can write two more inequalities based on this: \[x \geq 0\] \[y \geq 0\] Now, we have our system of linear inequalities for this problem: \[x + y \leq 12\] \[x \geq 0\] \[y \geq 0\]

Step 2: Graph the feasible region

To graph the feasible region, we will first plot the line \(x+y=12\). This line will divide the graph into two regions. The region where \(x+y \leq 12\) will be our feasible region. Now, the feasible region will be a triangle with vertices at \((0,0)\), \((0,12)\), and \((12,0)\). These are the intersection points of the three inequalities.

Step 3: Find three points inside the feasible region and discuss their meanings

Let's choose the following three points inside the feasible region: 1. \((2,6)\): In this case, the machine fills large bags for 2 hours and small bags for 6 hours, for a total of 8 hours of work in a day. This distribution allows for production for 2 hours and 6 minutes for the large and small bags offered. 2. \((4,4)\): Here, the machine fills large bags for 4 hours per day and small bags for 4 hours per day as well. The machine works for a total of 8 hours, which is within the constraint of 12 working hours. 3. \((6,2)\): The machine fills large bags for 6 hours and small bags for 2 hours in this case. The total working hours are 8, which is within the constraint. However, this distribution prioritizes filling large bags over small bags.

Step 4: Find one point outside the feasible region and discuss its meaning

Let's choose the point \((10,10)\), which is outside the feasible region. In this case, the machine fills large bags for 10 hours and small bags for 10 hours every day. However, this would mean that the machine needs to work for 20 hours, which is more than the allowed working hours of 12 hours per day.

Key Concepts

Feasible Region

The feasible region represents all possible combinations of variables that satisfy a given set of inequalities. In the context of our exercise, it shows all possible ways the machine can allocate its time between filling large and small bags, without exceeding its maximum run time of 12 hours per day.

To determine the feasible region on a graph, we plot the system of inequalities, which are:

• \(x + y \leq 12\): The total number of hours cannot exceed 12.
• \(x \geq 0\): The machine can't run for negative hours on large bags.
• \(y \geq 0\): The machine can't run for negative hours on small bags.

These inequalities form a triangular region on the graph. Points within this region, such as \((0,0)\), \((0,12)\), and \((12,0)\), indicate feasible time distributions. Any point in the feasible region, such as \((4,4)\), represents a valid way to use the available 12 hours, confirming its utility in operational scheduling.

Graphing Inequalities

Graphing inequalities is a fundamental concept used to visually represent solutions to inequality constraints, like those in our system.

When graphing the inequality \(x + y \leq 12\), you first plot the boundary line \(x + y = 12\). This line represents the maximum boundaries of when the machine could theoretically operate right up to its limit.

• Start by finding intercepts: where the line cuts the x-axis (\((12, 0)\)) and the y-axis (\((0, 12)\)).
• Use these intercepts to draw the line.
• Since \(x + y\) needs to be less than or equal to 12, you'll shade below this line.

Additionally, the lines \(x \geq 0\) and \(y \geq 0\) ensure we only consider non-negative values, as these restrict the graph to the first quadrant. Together, these aspects help pinpoint the feasible region.

System of Inequalities

A system of inequalities involves two or more inequalities that are considered simultaneously. This is crucial for real-world scenarios where multiple conditions need to be fulfilled at the same time.

In the exercise, the system of inequalities was formulated to model constraints on operational hours. This system includes:

• \(x + y \leq 12\): Ensures the machine doesn't exceed its maximum operation time.
• \(x \geq 0\) and \(y \geq 0\): Ensure practical scenarios where time values are non-negative.

By interpreting this system, we can explore all permissible solutions describing how much time the machine can spend on filling large or small bags within a day. By solving or graphing these inequality constraints together, you can determine feasible solutions that meet all constraints and optimize operations effectively.