Finding Vertices Of A Feasible Region In Linear Programming

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Hey guys! Today, we're diving deep into the fascinating world of linear programming, where we'll tackle a classic problem: finding the vertices of a feasible region defined by a set of constraints. Don't worry if that sounds intimidating – we'll break it down into easy-to-understand steps. We'll use a blend of graphical and algebraic methods to solve the problem, and by the end, you'll be a pro at identifying these crucial points. So, grab your pencils, and let's get started!

Defining the Feasible Region: Our Starting Point

First, let's clearly state the constraints we're working with. These inequalities define the boundaries of our feasible region, which is the set of all points (x, y) that satisfy all the constraints simultaneously. Here they are:

  1. 4x + 3y ≤ 12
  2. 2x + 6y ≤ 15
  3. x ≥ 0
  4. y ≥ 0

The last two constraints, x ≥ 0 and y ≥ 0, tell us that we're only interested in the first quadrant of the coordinate plane (where both x and y are non-negative). This simplifies our task a bit, as we don't have to worry about the other quadrants.

Now, let's discuss why finding the vertices of this feasible region is so important. In linear programming, we often want to maximize or minimize an objective function (like profit or cost) subject to these constraints. A fundamental theorem in linear programming states that the optimal solution (the maximum or minimum value) will always occur at one of the vertices of the feasible region. Therefore, by identifying these vertices, we narrow down our search for the optimal solution to a finite set of points – a much more manageable task than searching the entire feasible region!

So, our goal is clear: we need to find the coordinates (x, y) of the vertices where the boundary lines of our feasible region intersect. Let's move on to the next step and graph these inequalities to visualize the region.

Graphing the Inequalities: Visualizing the Feasible Region

To visualize the feasible region, we need to graph each inequality on the coordinate plane. The best way to do this is to first treat each inequality as an equation and graph the corresponding line. Then, we'll determine which side of the line satisfies the inequality.

Let's start with the first inequality:

4x + 3y ≤ 12

To graph the line 4x + 3y = 12, we can find two points on the line. A simple way to do this is to set x = 0 and solve for y, and then set y = 0 and solve for x:

  • If x = 0, then 3y = 12, so y = 4. This gives us the point (0, 4).
  • If y = 0, then 4x = 12, so x = 3. This gives us the point (3, 0).

Plot these two points on the coordinate plane and draw a line through them. This line represents the equation 4x + 3y = 12. Now, we need to determine which side of the line represents the inequality 4x + 3y ≤ 12. To do this, we can pick a test point (not on the line) and see if it satisfies the inequality. A good choice is the origin, (0, 0):

  • 4(0) + 3(0) = 0 ≤ 12

Since the origin satisfies the inequality, the region below the line 4x + 3y = 12 is the region that satisfies 4x + 3y ≤ 12. We'll shade this region to indicate that it's part of our feasible region.

Now, let's move on to the second inequality:

2x + 6y ≤ 15

Similarly, we graph the line 2x + 6y = 15 by finding two points:

  • If x = 0, then 6y = 15, so y = 2.5. This gives us the point (0, 2.5).
  • If y = 0, then 2x = 15, so x = 7.5. This gives us the point (7.5, 0).

Plot these points and draw the line. Again, we use the origin as a test point:

  • 2(0) + 6(0) = 0 ≤ 15

The origin satisfies the inequality, so we shade the region below the line 2x + 6y = 15.

Finally, the inequalities x ≥ 0 and y ≥ 0 simply restrict our feasible region to the first quadrant. This means we're only interested in the area where x and y are both positive or zero.

After graphing all the lines and shading the appropriate regions, the feasible region is the area that is shaded by all the inequalities. It's the intersection of all the shaded regions. This region is a polygon, and its vertices are the points where the boundary lines intersect. These are the points we're after!

Identifying the Vertices: Where the Lines Meet

Now that we have a visual representation of the feasible region, we can identify the vertices. These are the corner points of the polygon formed by the intersecting lines. By carefully examining the graph, we can see that there are four vertices. Some vertices can be read directly from the graph, while others require a bit of algebra to find their exact coordinates.

  1. The Origin: This is the easiest vertex to identify. It's the point where the x-axis and y-axis intersect, which is (0, 0). This point clearly satisfies all four constraints.
  2. The y-intercept of 2x + 6y = 15: This vertex lies on the y-axis, where x = 0. We already found this point when graphing the line 2x + 6y = 15. It's the point (0, 2.5). To double-check, we can substitute these coordinates into the inequalities:
    • 4(0) + 3(2.5) = 7.5 ≤ 12 (satisfied)
    • 2(0) + 6(2.5) = 15 ≤ 15 (satisfied)
    • 0 ≥ 0 (satisfied)
    • 2.5 ≥ 0 (satisfied)
  3. The x-intercept of 4x + 3y = 12: This vertex lies on the x-axis, where y = 0. We found this point when graphing the line 4x + 3y = 12. It's the point (3, 0). Let's check if it satisfies the constraints:
    • 4(3) + 3(0) = 12 ≤ 12 (satisfied)
    • 2(3) + 6(0) = 6 ≤ 15 (satisfied)
    • 3 ≥ 0 (satisfied)
    • 0 ≥ 0 (satisfied)
  4. The Intersection of 4x + 3y = 12 and 2x + 6y = 15: This is the trickiest vertex to find because it's the point where the two main lines intersect. We need to use algebra to solve this system of equations. There are a couple of ways to do this – substitution or elimination. Let's use elimination.

To eliminate x, we can multiply the second equation by -2:

  • 4x + 3y = 12
  • -4x - 12y = -30

Now, add the two equations together:

  • -9y = -18

Divide both sides by -9:

  • y = 2

Now that we have the value of y, we can substitute it back into either of the original equations to solve for x. Let's use the first equation:

  • 4x + 3(2) = 12
  • 4x + 6 = 12
  • 4x = 6
  • x = 1.5

So, the intersection point is (1.5, 2). Let's verify that this point satisfies all the constraints:

*   ***4(1.5) + 3(2) = 6 + 6 = 12 ≤ 12*** (satisfied)
*   ***2(1.5) + 6(2) = 3 + 12 = 15 ≤ 15*** (satisfied)
*   ***1.5 ≥ 0*** (satisfied)
*   ***2 ≥ 0*** (satisfied)

Great! We've found all four vertices of the feasible region.

The Vertices of the Feasible Region: Our Final Answer

After carefully analyzing the graph and performing the necessary algebra, we've successfully identified all the vertices of the feasible region. These are the corner points that define the boundaries of our solution space. To summarize, the vertices are:

  • (0, 0)
  • (0, 2.5)
  • (1.5, 2)
  • (3, 0)

Therefore, the answer to the question