Solving Linear Inequalities & Optimizing Functions

by TextBrain Team 51 views

Hey guys! Let's dive into a fun math problem involving linear inequalities and how to find the maximum or minimum value of a function within a specific region. We'll be using the following inequalities:

x + y \le 6
2x + y \ge 6
y \ge x
f(x,y) = 6x + 5y

We'll explore how to graph these inequalities, identify the feasible region, and then optimize (maximize or minimize) the function f(x, y) within that region. This is a classic example of linear programming, a powerful technique used in various fields to make the best decisions under certain constraints. Get ready to brush up on your algebra skills, because we're about to have some fun! We'll break down each step, making sure it's easy to follow along. Let's begin by understanding each inequality individually.

Understanding the Inequalities

First, let's break down these inequalities, piece by piece, so we can understand them better. To graph these, we'll treat each inequality as an equation first and find the points where they intersect the x and y axes. This helps us draw the lines that define our regions. So, let's start with x + y ≤ 6. The corresponding equation is x + y = 6. To graph this, let's find the intercepts. When x = 0, y = 6, and when y = 0, x = 6. This gives us the points (0, 6) and (6, 0). We draw a straight line through these points. Because the inequality is less than or equal to, the region we're interested in is below this line, including the line itself. It is going to be a solid line, not a dashed one. Next is 2x + y ≥ 6. As an equation, it is 2x + y = 6. When x = 0, y = 6, and when y = 0, x = 3. That yields the points (0, 6) and (3, 0). We'll draw a line through these points, again solid, because of the 'or equal to' part. The region here is above this line because of the 'greater than or equal to' sign. Finally, we have y ≥ x. In equation form, this is y = x. This is a simple line that passes through the origin (0,0), with a slope of 1. The region we are interested in lies above this line, because of the greater than or equal to sign. Also, because it is greater or equal, it is a solid line. Understanding each of these inequalities separately sets the stage for finding the feasible region.

Now that we have graphed each inequality and figured out which side of the line represents the solution set, we must identify the feasible region, the area where all inequalities are satisfied simultaneously. This is the area where all the shaded regions overlap. This intersection is crucial because it represents all the points (x, y) that satisfy all the original constraints. The feasible region is the foundation for solving the optimization problem because our solutions must come from this area. Any solution outside this area does not meet all the conditions. Remember, this region could be a bounded polygon, an unbounded region, or, in some cases, there might be no feasible region at all if the constraints are contradictory. The feasible region helps to visualize the potential solutions. It is not just about graphing; it is about understanding the relationships between the inequalities and their effect on the possible values of x and y. In this case, our feasible region will be a polygon.

Finding the Feasible Region

Alright, now let's pinpoint that feasible region. This is the area on the graph where all the inequalities are true. It's like finding the intersection of all the shaded areas we identified in the previous step. The feasible region is like the playing field within which our solution must lie. The boundaries of the feasible region are formed by the lines we graphed earlier: x + y = 6, 2x + y = 6, and y = x. Where these lines intersect is very important. The points where these lines cross each other are called vertices. The vertices are where the edges of the feasible region meet. This is where the fun begins, as the solution to the optimization problem (maximizing or minimizing our function) usually lies at one of these vertices. To identify these vertices, we must solve systems of equations where the lines intersect. For example, where does x + y = 6 intersect 2x + y = 6? Solving these systems is a straightforward process: you can use substitution, elimination, or any method you like. Find all the vertices by solving these systems of equations! The intersection of x + y = 6 and y = x gives us one vertex; the intersection of 2x + y = 6 and y = x gives us another; and the intersection of x + y = 6 and 2x + y = 6 gives us the last one. These are the corners of our playing field, the essential points we need to evaluate in our next step.

Remember, the constraints are not just lines on a graph; they define the scope within which our solution must exist. It is all about visualizing the constraints to guide your decision-making, and it helps in grasping complex relationships, making it a crucial tool for optimization. It is an iterative process, meaning you might refine your approach as you graph and analyze the inequalities. Let's move on to identify the vertices.

Identifying the Vertices

Now that we've understood the feasible region, let's find the vertices. These are the corner points of our feasible region, and they're super important. Vertices are the points where the boundary lines of the feasible region intersect. In our case, we need to solve the systems of equations formed by pairs of the lines we graphed: x + y = 6, 2x + y = 6, and y = x. Let's find the intersection points:

  1. Intersection of x + y = 6 and y = x: Substituting y = x into the first equation, we get x + x = 6, or 2x = 6. This gives us x = 3. Since y = x, we have y = 3. Thus, the vertex is (3, 3).
  2. Intersection of 2x + y = 6 and y = x: Substituting y = x, we get 2x + x = 6, or 3x = 6. Thus, x = 2. So, y = 2. The vertex is (2, 2).
  3. Intersection of x + y = 6 and 2x + y = 6: We can solve this using elimination. Subtracting the first equation from the second, we get (2x + y) - (x + y) = 6 - 6, which simplifies to x = 0. Substituting x = 0 into x + y = 6, we get y = 6. The vertex is (0, 6).

So, the vertices of our feasible region are (3, 3), (2, 2), and (0, 6). These vertices are our candidates for the maximum or minimum value of f(x, y). Now that we have the vertices, we can plug these points into f(x, y) to find the optimal solution.

Optimizing the Function

Finally, let's optimize our function, f(x, y) = 6x + 5y. We will evaluate this function at each of the vertices we found to determine the maximum and minimum values within the feasible region. This is the culmination of all our work. We'll plug in the x and y values from each vertex into the function f(x, y) and calculate the result. By comparing these results, we can pinpoint the optimal solution. Let's evaluate:

  • At vertex (3, 3): f(3, 3) = 6(3) + 5(3) = 18 + 15 = 33.
  • At vertex (2, 2): f(2, 2) = 6(2) + 5(2) = 12 + 10 = 22.
  • At vertex (0, 6): f(0, 6) = 6(0) + 5(6) = 0 + 30 = 30.

By evaluating the function at each vertex, we can determine both the maximum and minimum values of f(x, y) within our feasible region. In this case, the maximum value of f(x, y) is 33, which occurs at the point (3, 3), and the minimum value is 22, which occurs at the point (2, 2). Therefore, we have successfully optimized our function under the given constraints.

Conclusion

So there you have it, guys! We've successfully navigated the world of linear inequalities and optimization. We learned how to graph inequalities, identify the feasible region, find vertices, and optimize a function. This approach is widely used in various fields, from business and economics to engineering and operations research, to maximize profits, minimize costs, or make the best possible decisions given certain limitations. Understanding these concepts provides a powerful tool for tackling real-world problems. This method is a fundamental part of linear programming, and understanding it opens doors to a multitude of practical applications.

Keep practicing these problems, and you'll become a pro in no time!