Minimize Z = 2x + 3y: A Simple Guide

by Henrik Larsen 37 views

Hey guys! Ever stumbled upon a problem that looks like a jumbled mess of equations and inequalities? Well, today we're diving into one of those – a classic optimization problem. Don't worry, it's not as scary as it looks! We're going to break down how to minimize the function Z = 2x + 3y, subject to some constraints. Think of it as a puzzle where we need to find the best possible solution while staying within the rules. Let's get started!

Understanding the Problem: Minimizing Z = 2x + 3y

So, what exactly are we trying to do here? At its core, this problem is about finding the smallest possible value for the expression Z = 2x + 3y. This expression is what we call our objective function. Think of it as the thing we want to make as small as we can. But, there's a catch! We can't just pick any values for x and y. We have to play by the rules, which are set by our constraints. These constraints are like fences that keep us within a certain area on a graph. In our case, we have three constraints: 20x + 60y ≥ 1200, 20x + 15y ≤ 480, and x, y ≥ 0. The last one, x, y ≥ 0, simply means that x and y can't be negative – we're only dealing with the positive quadrant of the graph.

Why is this useful? you might ask. Well, these kinds of problems pop up all over the place in the real world! Imagine you're a business owner trying to minimize costs while still meeting production demands, or a nutritionist trying to create a meal plan that meets dietary requirements with the fewest calories. These are all situations where optimization techniques come in handy. Linear programming, which is what we're using here, is a powerful tool for solving these types of problems. So, understanding how to tackle this problem is a valuable skill to have.

To really grasp this, let's break down each component. Our objective function, Z = 2x + 3y, is a linear equation. This means that when we graph it, it will be a straight line. The constraints are also linear inequalities, which means they will define regions on the graph. The solution to our problem lies within the feasible region – the area where all the constraints overlap. We're looking for the point within this region that gives us the smallest possible value for Z. It's like searching for the lowest point in a valley! By understanding the problem in this way, we can approach it systematically and find the optimal solution. So, let's dive into the next step: graphing these constraints.

Graphing the Constraints: Visualizing the Feasible Region

Alright, so now we know what we're trying to minimize and the rules we have to follow. The next step is to visualize these constraints. Graphing the inequalities helps us see the feasible region, which is the area where all the constraints are satisfied. Think of it as the playing field where our solution lies. To graph these inequalities, we'll first treat them as equations and then determine which side of the line represents the solution region.

Let's start with the first constraint: 20x + 60y ≥ 1200. To graph this, we'll first turn it into an equation: 20x + 60y = 1200. Now, we need to find two points on this 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, we have 60y = 1200, so y = 20. This gives us the point (0, 20). If y = 0, we have 20x = 1200, so x = 60. This gives us the point (60, 0). We can plot these two points on a graph and draw a line through them. But, here's the key: since our original constraint was 20x + 60y ≥ 1200, we need to figure out which side of the line represents the solutions. To do this, we can test a point that's not on the line, like (0, 0). Plugging in (0, 0) into the inequality, we get 20(0) + 60(0) ≥ 1200, which simplifies to 0 ≥ 1200. This is false, so the region that includes (0, 0) is not part of our solution. We shade the other side of the line to represent the solution region for this constraint.

Now, let's tackle the second constraint: 20x + 15y ≤ 480. We'll follow the same process. First, we turn it into an equation: 20x + 15y = 480. If x = 0, we have 15y = 480, so y = 32. This gives us the point (0, 32). If y = 0, we have 20x = 480, so x = 24. This gives us the point (24, 0). We plot these points and draw a line. This time, we test the point (0, 0) in the inequality 20x + 15y ≤ 480. Plugging in (0, 0), we get 0 ≤ 480, which is true. So, the region that includes (0, 0) is part of our solution, and we shade that side of the line.

Finally, we have the constraints x, y ≥ 0. This simply means we're only interested in the first quadrant of the graph, where both x and y are non-negative. So, we can imagine lines along the x and y axes, and we're only looking at the region to the right of the y-axis and above the x-axis.

When we overlay all these shaded regions, the area where they all overlap is our feasible region. This is the set of all possible solutions that satisfy all the constraints. It might look like a polygon on the graph. The corners of this polygon are particularly important because they are the potential locations of our optimal solution. So, the next step is to identify these corner points.

Identifying Corner Points: Where the Magic Happens

Okay, we've graphed our constraints and found the feasible region – the playground for our solutions. Now, we need to pinpoint the corner points of this region. These points are crucial because the optimal solution (the minimum value of Z in our case) will always occur at one of these corners. Think of it like this: if you're trying to find the lowest point in a valley, you're likely to find it at the bottom of a slope or in a corner where different slopes meet. The same principle applies here.

The corner points are the intersections of the lines that define our feasible region. We can find these points by solving the systems of equations formed by the intersecting lines. Let's look at our constraints again: 20x + 60y = 1200, 20x + 15y = 480, x = 0, and y = 0. We've already used x = 0 and y = 0 to find points on our lines, and they also represent the axes, so they're likely to be part of our corner points.

One corner point is easy to spot: it's the intersection of the x and y axes, which is the origin (0, 0). However, this point might not be within our feasible region because it might not satisfy all the constraints. We'll check this later.

Another corner point will be the intersection of the line 20x + 60y = 1200 with one of the axes. We already found that this line intersects the y-axis at (0, 20). This is a potential corner point. Similarly, the line 20x + 15y = 480 intersects the y-axis at (0, 32) and the x-axis at (24, 0). So, (0, 32) and (24, 0) are also potential corner points.

But, we're not done yet! We also need to find the intersection point of the two lines: 20x + 60y = 1200 and 20x + 15y = 480. This is where we need to solve a system of equations. There are several ways to do this, like substitution or elimination. Let's use elimination. We can subtract the second equation from the first to eliminate x: (20x + 60y) - (20x + 15y) = 1200 - 480, which simplifies to 45y = 720. Dividing both sides by 45, we get y = 16. Now, we can plug this value of y back into either equation to solve for x. Let's use the second equation: 20x + 15(16) = 480, which simplifies to 20x + 240 = 480. Subtracting 240 from both sides gives us 20x = 240, and dividing by 20 gives us x = 12. So, the intersection point of these two lines is (12, 16).

Now, we have a set of potential corner points: (0, 0), (0, 20), (0, 32), (24, 0), and (12, 16). To be absolutely sure, we need to check if all these points actually lie within our feasible region by plugging them into our original constraints. Any point that doesn't satisfy all constraints is not a valid corner point. Once we've identified our true corner points, we're ready for the final step: evaluating our objective function at these points.

Evaluating the Objective Function: Finding the Minimum Z

Alright, we've done the hard work of graphing the constraints and finding the corner points of our feasible region. Now comes the moment of truth: evaluating our objective function Z = 2x + 3y at these corner points. This will tell us which point gives us the smallest value for Z, which is what we're trying to minimize. It's like the final lap in a race, where we see which corner point crosses the finish line with the lowest Z score.

Let's take each corner point we identified and plug its x and y coordinates into our objective function:

  • Point (0, 0): Z = 2(0) + 3(0) = 0
  • Point (0, 20): Z = 2(0) + 3(20) = 60
  • Point (24, 0): Z = 2(24) + 3(0) = 48
  • Point (12, 16): Z = 2(12) + 3(16) = 24 + 48 = 72

Now, let's compare the values of Z we calculated. We have Z = 0, Z = 60, Z = 48, and Z = 72. The smallest value is Z = 0, which occurs at the point (0, 0). However, we need to remember our constraints. While (0, 0) gives us the smallest value for Z, we need to check if it actually satisfies all our constraints. Let's plug (0, 0) into our constraints:

  • 20(0) + 60(0) ≥ 1200 => 0 ≥ 1200 (False)
  • 20(0) + 15(0) ≤ 480 => 0 ≤ 480 (True)
  • 0 ≥ 0 (True)
  • 0 ≥ 0 (True)

Since the point (0, 0) does not satisfy the first constraint (0 ≥ 1200 is false), it's not a feasible solution. This is a crucial step – we can't just pick the point that gives us the smallest Z value; it has to be within our feasible region! So, (0,0) is not our solution.

Let's look at the other Z values. The next smallest is Z = 48, which occurs at the point (24, 0). Let's check if this point is feasible:

  • 20(24) + 60(0) ≥ 1200 => 480 ≥ 1200 (False)

Since (24, 0) also fails to satisfy the first constraint, it's not a feasible solution either. We move on to the next smallest value, Z = 60, which occurs at the point (0, 20). Let's check feasibility:

  • 20(0) + 60(20) ≥ 1200 => 1200 ≥ 1200 (True)
  • 20(0) + 15(20) ≤ 480 => 300 ≤ 480 (True)
  • 0 ≥ 0 (True)
  • 20 ≥ 0 (True)

The point (0, 20) satisfies all the constraints! This means it's a feasible solution, and it gives us Z = 60. Now, let's check the last point,(12, 16) with Z = 72:

  • 20(12) + 60(16) ≥ 1200 => 1200 ≥ 1200 (True)
  • 20(12) + 15(16) ≤ 480 => 480 ≤ 480 (True)
  • 12 ≥ 0 (True)
  • 16 ≥ 0 (True)

Since the point (12, 16) satisfies all the constraints and the other point (0,20) gives us the smallest Z value, we have our solution!

The Optimal Solution: Our Final Answer

We've reached the end of our journey! We started with a seemingly complex problem, but by breaking it down step by step, we've found the optimal solution. Let's recap what we did:

  1. We understood the problem: We identified our objective function (Z = 2x + 3y) and our constraints (20x + 60y ≥ 1200, 20x + 15y ≤ 480, x, y ≥ 0).
  2. We graphed the constraints: We turned the inequalities into equations, plotted the lines, and shaded the feasible region.
  3. We identified corner points: We found the intersections of the lines that defined our feasible region by solving systems of equations.
  4. We evaluated the objective function: We plugged the coordinates of each corner point into Z = 2x + 3y and compared the results.
  5. We verified that the points meet the constrains to make sure our results are correct.

After all this, we found that the minimum value of Z is 60, which occurs at the point (0, 20). This means that the optimal solution to our problem is x = 0 and y = 20.

So, there you have it! We've successfully minimized Z subject to our given constraints. These optimization problems might seem tough at first, but with a systematic approach and a little bit of graphing, you can conquer them. Keep practicing, and you'll become a pro at finding optimal solutions in no time!