Graphing Linear Inequalities: Boundedness & Corner Points

Hey Plastik Magazine readers! Today, we're diving into the fascinating world of linear inequalities and how to graph them. We'll specifically tackle the system:

{
  x ≥ 0
  y ≥ 0
  5x + y ≤ 30
  x + 5y ≤ 30
}

We'll break down how to graph these inequalities, figure out if the resulting graph is bounded or unbounded, and pinpoint those crucial corner points. So, grab your graph paper (or your favorite digital graphing tool) and let's get started!

Graphing the System of Linear Inequalities

So you want to learn how to graph a system of linear inequalities? Graphing linear inequalities might seem a little daunting at first, but trust me, it's totally manageable once you understand the steps. It all boils down to visualizing the solution set, which is the region where all the inequalities are satisfied simultaneously. This is super useful in various applications, like optimization problems and resource allocation. Okay, let's break down how we can graph the system of linear inequalities. First we need to graph each inequality individually. Remember that a linear inequality represents a half-plane on the coordinate plane. The boundary line of this half-plane is the line you get by replacing the inequality sign (≤, ≥, <, >) with an equal sign (=). This boundary line is crucial, because it separates the solutions from the non-solutions. We have four inequalities here: x ≥ 0, y ≥ 0, 5x + y ≤ 30, and x + 5y ≤ 30. Let's take them one by one. The inequalities x ≥ 0 and y ≥ 0 are straightforward. x ≥ 0 simply means we're looking at the region to the right of the y-axis, including the y-axis itself. Similarly, y ≥ 0 means we're looking at the region above the x-axis, including the x-axis. So, right off the bat, we know our solution will lie in the first quadrant (where both x and y are non-negative). Now let's tackle the more complex inequalities: 5x + y ≤ 30 and x + 5y ≤ 30. For 5x + y ≤ 30, we first graph the line 5x + y = 30. To do this, we can find two points on the line. A simple way is to find the intercepts. When x = 0, we have y = 30, giving us the point (0, 30). When y = 0, we have 5x = 30, so x = 6, giving us the point (6, 0). Plot these two points and draw a line through them. This is the boundary line. But we're not done yet! We need to figure out which side of the line represents the solution to the inequality 5x + y ≤ 30. We can do this by picking a test point, a point that is not on the line. The easiest one to use is often the origin, (0, 0). Substitute x = 0 and y = 0 into the inequality: 5(0) + 0 ≤ 30. This simplifies to 0 ≤ 30, which is true! Since the origin satisfies the inequality, the region containing the origin is the solution region. Shade this region. For x + 5y ≤ 30, we follow the same process. First, graph the line x + 5y = 30. When x = 0, we have 5y = 30, so y = 6, giving us the point (0, 6). When y = 0, we have x = 30, giving us the point (30, 0). Plot these points and draw the line. Again, we use a test point to determine which side of the line to shade. Let's use the origin (0, 0). Substitute into the inequality: 0 + 5(0) ≤ 30. This simplifies to 0 ≤ 30, which is true. So, we shade the region containing the origin. Now, here's the key: the solution to the system of inequalities is the region where all the shaded regions overlap. This is the region that satisfies all four inequalities simultaneously. It's like the sweet spot where everything works! You should end up with a four-sided polygon in the first quadrant. This polygon is the graphical representation of the solution to your system of inequalities. Congrats, you've graphed the system! But we still need to figure out if it's bounded or unbounded and find those corner points. Keep reading! By following these steps carefully, you'll be graphing systems of linear inequalities like a pro in no time. Remember, practice makes perfect, so try graphing a few more systems on your own to really solidify your understanding.

Visualizing the Solution Region

Now, when we visualize the solution region, think of it as the area on the graph where all the conditions of our inequalities are met. It's like a Venn diagram, but with shaded areas! Remember those four inequalities we talked about? x ≥ 0, y ≥ 0, 5x + y ≤ 30, and x + 5y ≤ 30? Each one creates a specific boundary on our graph, and the solution region is where all those boundaries overlap. It's kind of like finding the common ground for all the inequalities. We already know that x ≥ 0 and y ≥ 0 restrict us to the first quadrant, which is the top-right corner of our graph where both x and y values are positive or zero. This is our starting point, our frame of reference. Now, let's bring in the other two inequalities. The lines 5x + y = 30 and x + 5y = 30 act as additional boundaries, cutting off parts of the first quadrant. Imagine each line as a fence, and the inequalities tell us which side of the fence we're allowed to be on. For 5x + y ≤ 30, we shaded the region below the line because the origin (0, 0) satisfied the inequality. This means all the points in that shaded region, and on the line itself, are solutions to this specific inequality. Similarly, for x + 5y ≤ 30, we shaded the region below its line as well, for the same reason. So, what's the solution region for the entire system? It's the area where the shading from all four inequalities overlaps. It's like the intersection of all the allowed zones. If you've graphed it correctly, you should see a four-sided polygon in the first quadrant. This polygon is your solution region! Every single point inside this polygon, and on its edges, represents a pair of (x, y) values that satisfy all four inequalities simultaneously. That's a pretty powerful visualization! This solution region is super important. It's not just a pretty shape on the graph; it represents all the possible solutions to our system of inequalities. This is where you can find the answers to real-world problems that can be modeled using inequalities, like maximizing profits or minimizing costs, subject to certain constraints. So, take a good look at your graph. Make sure you understand why that particular polygon is the solution region. Imagine picking any point inside it – it should satisfy all four inequalities. And if you pick a point outside the polygon, it will fail at least one of them. That's the beauty of visualizing the solution region – it gives you a clear picture of all the possibilities. Now that we've got our solution region visualized, let's move on to the next question: Is this region bounded or unbounded? This will tell us something important about the nature of our solutions. Let's dive in!

Determining Bounded or Unbounded

Okay, guys, let's figure out if our graphed region is bounded or unbounded. This is actually a pretty straightforward concept. Think of it like this: can you draw a circle big enough to completely enclose the solution region? If you can, then it's bounded. If the solution region stretches off infinitely in some direction, then it's unbounded. In our case, we have the inequalities x ≥ 0 and y ≥ 0, which already restrict us to the first quadrant. This means our region won't stretch infinitely to the left or downwards. But what about upwards and to the right? That's where the other two inequalities come in: 5x + y ≤ 30 and x + 5y ≤ 30. These inequalities act as further constraints, essentially