Minimize Z = 8X1 + 3X2: A Step-by-Step Guide

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Hey guys! Today, we're diving into a classic optimization problem: minimizing a function with constraints. Specifically, we'll be tackling the problem of minimizing Z = 8X1 + 3X2 subject to a set of inequalities. This is a common type of problem in linear programming, and it has applications in various fields, from business to engineering. So, let's break it down and make sure you understand every step.

Understanding the Problem

Before we jump into the solution, let's make sure we all understand what we're trying to achieve. We have an objective function, Z = 8X1 + 3X2, which we want to make as small as possible. Think of Z as the cost we want to reduce. X1 and X2 are our decision variables – the values we can control to minimize Z. But, there are rules! We can't just pick any values for X1 and X2. We have constraints, which are inequalities that limit our choices. These constraints represent real-world limitations, like resource availability or production requirements. Our goal is to find the values of X1 and X2 that minimize Z while still satisfying all the constraints. Sounds like a fun puzzle, right? We're going to solve this using a graphical approach, which is super helpful for visualizing what's going on.

Step 1: Graphing the Constraints

Okay, the first thing we need to do is graph each of our constraints. This will help us visualize the feasible region, which is the set of all possible solutions that satisfy all the constraints simultaneously. Think of the feasible region as our playground – the only area where we're allowed to play. Let's look at our constraints:

  1. 12X1 + 9X2 ≥ 6
  2. 3X1 + 6X2 ≥ 12
  3. 6X1 + X2 ≥ 6
  4. X1, X2 ≥ 0 (These are non-negativity constraints, meaning X1 and X2 cannot be negative. This makes sense in many real-world scenarios, like you can't produce a negative amount of a product).

To graph each inequality, we'll first treat it as an equation and plot the line. Then, we'll determine which side of the line represents the solution to the inequality. Remember high school algebra? It's coming back to help us now!

Constraint 1: 12X1 + 9X2 ≥ 6

First, let's turn this into an equation: 12X1 + 9X2 = 6. To graph this line, we can find two points on the line. A simple way to do this is to set X1 to 0 and solve for X2, then set X2 to 0 and solve for X1.

  • If X1 = 0, then 9X2 = 6, so X2 = 6/9 = 2/3. Our first point is (0, 2/3).
  • If X2 = 0, then 12X1 = 6, so X1 = 6/12 = 1/2. Our second point is (1/2, 0).

Plot these points (0, 2/3) and (1/2, 0) on a graph and draw a line through them. Now, we need to figure out which side of the line satisfies the inequality 12X1 + 9X2 ≥ 6. To do this, we can test a point that's not on the line. The easiest point is often (0, 0). Plug it into the inequality:

12(0) + 9(0) ≥ 6 => 0 ≥ 6

This is false! Since (0, 0) does not satisfy the inequality, we shade the region away from (0, 0). This means we shade the region above and to the right of the line. This shaded area represents all the points that satisfy the first constraint.

Constraint 2: 3X1 + 6X2 ≥ 12

Let's do the same for the second constraint. First, the equation: 3X1 + 6X2 = 12.

  • If X1 = 0, then 6X2 = 12, so X2 = 2. Point: (0, 2).
  • If X2 = 0, then 3X1 = 12, so X1 = 4. Point: (4, 0).

Plot the points (0, 2) and (4, 0) and draw the line. Now, test the point (0, 0):

3(0) + 6(0) ≥ 12 => 0 ≥ 12

Again, this is false. So, we shade the region away from (0, 0), which is the region above and to the right of the line.

Constraint 3: 6X1 + X2 ≥ 6

Let's tackle the third constraint. The equation is 6X1 + X2 = 6.

  • If X1 = 0, then X2 = 6. Point: (0, 6).
  • If X2 = 0, then 6X1 = 6, so X1 = 1. Point: (1, 0).

Plot (0, 6) and (1, 0) and draw the line. Test (0, 0):

6(0) + (0) ≥ 6 => 0 ≥ 6

This is also false! Shade the region away from (0, 0), above and to the right of the line.

Constraint 4: X1, X2 ≥ 0

Finally, the non-negativity constraints X1, X2 ≥ 0 simply mean that we are only concerned with the first quadrant of the graph (where both X1 and X2 are positive or zero). This makes sense because we can't have negative production or negative resources, right?

Step 2: Identifying the Feasible Region

Now comes the fun part – putting it all together! The feasible region is the area on the graph where all the shaded regions from each constraint overlap. It's the area that satisfies all the inequalities simultaneously. Grab a different colored pen or highlighter and shade the area where all the individual shaded regions intersect. You'll likely end up with an unbounded region in the first quadrant, meaning it extends infinitely in some direction.

The feasible region will be a polygon (or an unbounded shape) formed by the intersection of the lines we graphed. The corners of this feasible region are called corner points, and they're crucial for finding the optimal solution. We need to identify these corner points. They are the points where the constraint lines intersect.

To find the exact coordinates of the corner points, we need to solve the system of equations for the intersecting lines. This might involve using substitution or elimination methods (remember those?).

Let's say we have corner points A, B, and C. We need to determine their coordinates by solving the pairs of equations that intersect at those points. This may seem a bit tedious, but it's a crucial step in finding the minimum value of our objective function.

Step 3: Evaluating the Objective Function at the Corner Points

Alright, we've found our playground (the feasible region) and identified the corners. Now, it's time to see where the lowest cost (minimum Z) hides. The key principle here is that the optimal solution (the minimum value of Z in our case) will always occur at one of the corner points of the feasible region. This is a fundamental concept in linear programming.

So, what we need to do is take our objective function, Z = 8X1 + 3X2, and plug in the coordinates of each corner point we identified in the previous step. This will give us the value of Z at each of these points. Then, we simply compare the values and choose the smallest one (since we're minimizing).

Let's say we have three corner points: A(X1a, X2a), B(X1b, X2b), and C(X1c, X2c). We'll calculate Z for each point:

  • Za = 8X1a + 3X2a
  • Zb = 8X1b + 3X2b
  • Zc = 8X1c + 3X2c

After calculating Za, Zb, and Zc, we compare these values. The smallest value will be the minimum value of Z, and the corresponding corner point will give us the values of X1 and X2 that minimize Z while satisfying all the constraints. This is our optimal solution!

Step 4: Interpreting the Results

We've crunched the numbers and found the minimum value of Z. Awesome! But what does it all mean? In the real world, X1 and X2 might represent quantities of products to produce, resources to allocate, or decisions to make. The minimum value of Z represents the lowest cost, the smallest amount of resource usage, or the optimal outcome we can achieve given the constraints.

For example, if X1 represents the number of units of product A to produce and X2 represents the number of units of product B, and Z represents the total production cost, then our solution tells us the optimal number of units of each product to produce to minimize the total cost while still meeting certain demands or limitations (represented by the constraints). Understanding the context of the problem is crucial for interpreting the solution and making informed decisions.

If the feasible region is unbounded, there's a possibility that the objective function can decrease indefinitely, meaning there's no minimum value. However, in many practical scenarios, the constraints will create a bounded feasible region, ensuring a finite optimal solution. It's always important to check for unboundedness, especially in theoretical problems.

Example Time!

Let's walk through a quick example to solidify our understanding. Suppose after graphing and finding the corner points, we have the following corner points for our feasible region:

  • A (0, 6)
  • B (1, 0)
  • C (4, 0)

Remember our objective function: Z = 8X1 + 3X2

Now, let's evaluate Z at each corner point:

  • At A (0, 6): Z = 8(0) + 3(6) = 18
  • At B (1, 0): Z = 8(1) + 3(0) = 8
  • At C (4, 0): Z = 8(4) + 3(0) = 32

Comparing the values, we see that the minimum value of Z is 8, which occurs at corner point B (1, 0). This means that to minimize Z, we should set X1 = 1 and X2 = 0.

Conclusion

So, there you have it! We've walked through the process of minimizing a linear objective function subject to constraints using the graphical method. We learned how to graph the constraints, identify the feasible region, find the corner points, evaluate the objective function, and interpret the results. This is a powerful technique that can be applied to a wide range of optimization problems. Keep practicing, and you'll become a pro at minimizing and maximizing in no time! Remember, understanding the problem, visualizing the constraints, and systematically evaluating the options are the keys to success. Now go out there and optimize the world, guys!