Minimize Z = 6x + 8y Graphically Step-by-Step Guide

by Marta Kowalska 52 views

Hey guys! Let's dive into a super useful mathematical technique: the graphical method for minimizing a linear objective function. In this article, we're going to break down how to minimize z = 6x + 8y subject to the constraints 40x + 10y ≥ 2400 and 10x + 15y ≥ 2100. Trust me, it's not as intimidating as it sounds! We'll go step-by-step, making it crystal clear and even a bit fun. So, grab your pencils and let’s get started!

Understanding the Problem

Before we jump into the graphical method, let's make sure we understand what we're trying to achieve. Our main goal is to minimize the objective function z = 6x + 8y. Think of z as the total cost we want to make as small as possible. The variables x and y are the decision variables, and they represent quantities that we can control. Now, there's a catch! We can't just pick any values for x and y. We have to follow certain rules, or constraints. These constraints are given as inequalities:

  1. 40x + 10y ≥ 2400
  2. 10x + 15y ≥ 2100

These inequalities tell us that the combinations of x and y must satisfy these minimum requirements. In simpler terms, we are looking for the smallest value of z that we can achieve while still meeting these conditions. It’s like trying to spend the least amount of money while still buying enough groceries to feed your family. The graphical method helps us visualize these constraints and find the optimal solution.

Why Use the Graphical Method?

You might be wondering, why bother with a graphical method? Well, it’s an incredibly intuitive way to solve linear programming problems, especially when we have two variables (like our x and y). It gives us a visual representation of the problem, which makes it easier to understand and solve. Instead of just crunching numbers, we can see the feasible region (the area where our solutions can exist) and identify the corner points. These corner points are crucial because the optimal solution (the minimum or maximum value of z) will always occur at one of these corners. The graphical method is not only educational but also practical for smaller problems. For larger, more complex problems with many variables, other methods like the simplex algorithm might be more efficient, but for our case, the graphical method is perfect!

Step-by-Step Graphical Solution

Okay, let's get into the nitty-gritty of solving this problem using the graphical method. We'll break it down into easy-to-follow steps. By the end of this section, you'll be a pro at plotting lines and finding feasible regions. So, let’s roll up our sleeves and dive in!

Step 1: Convert Inequalities to Equations

The first thing we need to do is turn our inequalities into equations. This is because equations are much easier to plot on a graph. So, we replace the "greater than or equal to" (≥) signs with equal signs (=). Our inequalities become:

  1. 40x + 10y = 2400
  2. 10x + 15y = 2100

These equations represent straight lines on our graph. Now, our mission is to plot these lines. Think of it as drawing the boundaries of our constraints. These boundaries will help us define the region where our solutions can exist. It's like setting the playing field for our problem. We need to find the right field (the feasible region) to win the game (minimize z).

Step 2: Find the Intercepts

To plot these lines, we need at least two points for each line. The easiest points to find are the intercepts – where the lines cross the x and y axes. Let's start with the first equation, 40x + 10y = 2400.

  • x-intercept: To find where the line crosses the x-axis, we set y = 0 and solve for x. So, 40x + 10(0) = 2400, which simplifies to 40x = 2400. Dividing both sides by 40, we get x = 60. This gives us the point (60, 0). Imagine this as the line touching the x-axis at the 60 mark. It's one of the anchor points for our line.
  • y-intercept: To find where the line crosses the y-axis, we set x = 0 and solve for y. So, 40(0) + 10y = 2400, which simplifies to 10y = 2400. Dividing both sides by 10, we get y = 240. This gives us the point (0, 240). Think of this as the line touching the y-axis at the 240 mark. Now we have another anchor point.

Now, let’s do the same for the second equation, 10x + 15y = 2100.

  • x-intercept: Set y = 0 and solve for x. So, 10x + 15(0) = 2100, which simplifies to 10x = 2100. Dividing both sides by 10, we get x = 210. This gives us the point (210, 0). This is our third anchor point.
  • y-intercept: Set x = 0 and solve for y. So, 10(0) + 15y = 2100, which simplifies to 15y = 2100. Dividing both sides by 15, we get y = 140. This gives us the point (0, 140). And finally, our fourth anchor point!

Now we have two points for each line. These points are like the cornerstones of our graph. We know exactly where our lines will pass through the axes. Next, we'll use these points to draw the lines on our graph. It’s like connecting the dots to reveal the shape of our solution space.

Step 3: Plot the Lines

Alright, we've got our intercepts, so it's time to put them on a graph! Grab a piece of graph paper (or use a digital graphing tool), and let's get to plotting. We're going to draw our coordinate axes – the x-axis and the y-axis. Make sure to scale your axes appropriately to fit the points we found in the previous step. Remember, our points are (60, 0), (0, 240) for the first line, and (210, 0), (0, 140) for the second line.

  • Plotting the first line (40x + 10y = 2400): Locate the point (60, 0) on the x-axis and the point (0, 240) on the y-axis. Now, carefully draw a straight line that passes through both of these points. This line represents all the possible combinations of x and y that satisfy the equation 40x + 10y = 2400. It’s like drawing a boundary – everything on this line meets our first condition exactly.
  • Plotting the second line (10x + 15y = 2100): Locate the point (210, 0) on the x-axis and the point (0, 140) on the y-axis. Draw another straight line that passes through both of these points. This line represents all the possible combinations of x and y that satisfy the equation 10x + 15y = 2100. This is our second boundary line.

Now, we have two lines on our graph. These lines divide the graph into different regions. But we're not interested in all the regions – we only care about the one that satisfies both of our original inequalities. Next, we'll figure out which side of each line represents the solutions to our inequalities. It’s like finding the right neighborhood to live in – one that meets all our requirements.

Step 4: Determine the Feasible Region

Now comes the crucial step: figuring out the feasible region. This is the area on the graph that satisfies all our constraints simultaneously. It's like finding the sweet spot where all our conditions are met. To do this, we'll look at our original inequalities and determine which side of each line represents the solutions.

  1. For the inequality 40x + 10y ≥ 2400: To figure out which side of the line satisfies this inequality, we can pick a test point. The easiest test point is usually the origin (0, 0). Plug x = 0 and y = 0 into the inequality: 40(0) + 10(0) ≥ 2400. This simplifies to 0 ≥ 2400, which is clearly false. Since (0, 0) does not satisfy the inequality, the feasible region is on the other side of the line. We shade the region above the line, away from the origin. This shaded area represents all the (x, y) pairs that satisfy the condition 40x + 10y ≥ 2400. Think of it as the territory governed by this constraint.
  2. For the inequality 10x + 15y ≥ 2100: Again, let's use the test point (0, 0). Plug x = 0 and y = 0 into the inequality: 10(0) + 15(0) ≥ 2100. This simplifies to 0 ≥ 2100, which is also false. So, the feasible region is on the other side of this line as well. We shade the region above this line, away from the origin. This shaded area represents all the (x, y) pairs that satisfy the condition 10x + 15y ≥ 2100. Another territory defined!

So, what's the feasible region? It's the area where the shaded regions from both inequalities overlap. This overlapping region represents all the points (x, y) that satisfy both inequalities. It's like the common ground where all our constraints agree. This region is often a polygon (a shape with straight sides), and its corners are particularly important. We call these corners corner points, and they're the key to finding our optimal solution. Next, we'll identify these corner points and see why they're so special.

Step 5: Identify the Corner Points

Now that we've found our feasible region, the next crucial step is to identify its corner points. These points are where the boundary lines intersect, and they're the vertices (corners) of our feasible region. Why are they so important? Well, in linear programming, the optimal solution (the minimum or maximum value of our objective function) will always occur at one of these corner points. It's like the hidden treasure is always buried at one of the corners of the island!

To find our corner points, we need to look at where our lines intersect. We have three types of corner points in this case:

  1. Intersection of the lines 40x + 10y = 2400 and 10x + 15y = 2100: This is where both constraints are active simultaneously. To find this point, we need to solve these two equations as a system. There are a couple of ways to do this: substitution or elimination. Let’s use elimination.
    • First, multiply the second equation by 4 to make the coefficients of x in both equations multiples of each other: 4 * (10x + 15y) = 4 * 2100, which gives us 40x + 60y = 8400.
    • Now, we have two equations:
      • 40x + 10y = 2400
      • 40x + 60y = 8400
    • Subtract the first equation from the second to eliminate x: (40x + 60y) - (40x + 10y) = 8400 - 2400, which simplifies to 50y = 6000.
    • Divide both sides by 50 to solve for y: y = 120.
    • Now that we have y, plug it back into one of the original equations to solve for x. Let’s use 40x + 10y = 2400: 40x + 10(120) = 2400, which simplifies to 40x + 1200 = 2400.
    • Subtract 1200 from both sides: 40x = 1200.
    • Divide both sides by 40 to solve for x: x = 30.
    • So, the intersection point is (30, 120). This is a key corner point where both our constraints are perfectly balanced.
  2. Intersection of the line 40x + 10y = 2400 and the x-axis: This is where y = 0. We already found this point when we calculated the x-intercept for the first line. It's (60, 0). This point represents the scenario where we're only using x, and we're at the minimum level required by the first constraint.
  3. Intersection of the line 10x + 15y = 2100 and the y-axis: This is where x = 0. We also found this point when we calculated the y-intercept for the second line. It's (0, 140). This point represents the scenario where we're only using y, and we're at the minimum level required by the second constraint.

So, our corner points are (30, 120), (60, 0), and (0, 140). These are the potential spots where our minimum cost z could be hiding. Next, we'll evaluate our objective function z at each of these points to see which one gives us the smallest value. It's like checking each corner of the treasure island to see where the loot is buried!

Step 6: Evaluate the Objective Function

Now comes the moment of truth! We've identified our corner points, and it's time to plug them into our objective function, z = 6x + 8y, to see which one gives us the smallest z value. Remember, we want to minimize z, so we're looking for the lowest number.

Let's evaluate z at each corner point:

  1. At point (30, 120):
    • z = 6(30) + 8(120)
    • z = 180 + 960
    • z = 1140 This is the value of z when we use a combination of x = 30 and y = 120. Keep this number in mind – it's a contender for the minimum.
  2. At point (60, 0):
    • z = 6(60) + 8(0)
    • z = 360 + 0
    • z = 360 Wow! This is much lower than our previous value. It seems like using more x and no y is significantly cheaper. But let’s not jump to conclusions yet – we have one more corner point to check.
  3. At point (0, 140):
    • z = 6(0) + 8(140)
    • z = 0 + 1120
    • z = 1120 This value is pretty high, almost as high as the first point we checked. It seems like using only y is not the most cost-effective strategy.

Okay, we've evaluated z at all three corner points. We have z = 1140 at (30, 120), z = 360 at (60, 0), and z = 1120 at (0, 140). It's clear that the smallest value of z is 360, which occurs at the point (60, 0). We've found our minimum! Next, we'll state our final answer and wrap up our graphical solution.

Conclusion: The Minimum Value

Alright, guys, we've reached the end of our journey through the graphical method! We started with a problem – minimizing z = 6x + 8y subject to some constraints – and we've successfully navigated our way to the solution. Let's recap what we've done and state our final answer.

We followed these steps:

  1. Converted inequalities to equations.
  2. Found the intercepts for each line.
  3. Plotted the lines on a graph.
  4. Determined the feasible region by shading the appropriate areas.
  5. Identified the corner points of the feasible region.
  6. Evaluated the objective function z at each corner point.

After all that work, we found that the minimum value of z is 360, and it occurs at the point (60, 0). This means that to minimize the cost, we should choose x = 60 and y = 0. In practical terms, this could mean producing 60 units of product X and 0 units of product Y, or any other scenario where x and y represent quantities. The key takeaway is that by using the graphical method, we were able to visually represent the problem, identify the constraints, and find the optimal solution.

The graphical method is a powerful tool for solving linear programming problems, especially when dealing with two variables. It provides an intuitive way to understand the problem and find the optimal solution. While more complex problems with many variables might require other methods like the simplex algorithm, the graphical method is an excellent starting point for understanding optimization. So, next time you encounter a minimization problem, remember the steps we've covered, and you'll be well-equipped to tackle it graphically. Keep practicing, and you'll become a pro in no time! Remember, math can be fun, especially when you can see the solutions right before your eyes! This comprehensive guide has hopefully made the graphical method much clearer and easier to understand. Happy solving!