Maximize P = 24x + 30y: A Step-by-Step Guide

Nov 11, 2025 by Admin 45 views

Hey guys! Today, we're diving into a classic optimization problem: how to maximize the value of a linear function, specifically P = 24x + 30y, when we have a bunch of constraints on our variables x and y. This type of problem pops up all over the place, from business to engineering, so understanding how to tackle it is super useful. We'll break it down step-by-step, making sure it's crystal clear. So, let's get started and figure out how to maximize P!

Understanding the Problem: Linear Programming

Before we jump into solving the problem, let's take a moment to understand what we're dealing with. This is a linear programming problem. Basically, we're trying to find the best possible value (either maximum or minimum) of a linear expression (our objective function), subject to a set of linear inequalities (our constraints). These constraints define a feasible region, which is the set of all possible (x, y) values that satisfy all the inequalities. The magic happens at the corners, or vertices, of this feasible region – that's where the maximum or minimum value of our objective function will always be found. Understanding this fundamental concept is super important because it guides our entire solution process. We're not just randomly plugging in numbers; we're strategically looking at the corners of a defined area.

Breaking Down the Components

Let's break down the key components of our problem:

Objective Function: P = 24x + 30y. This is the expression we want to maximize. Think of it as our target – the thing we're trying to make as big as possible.
Constraints:
- x + y ≤ 5
- x - y ≥ -1
- x ≥ 0
- y ≥ 0

These inequalities define the boundaries within which our solution must lie. They're like the rules of the game. We can't just pick any x and y; they have to fit within these constraints. The last two constraints, x ≥ 0 and y ≥ 0, simply mean that we're only working in the first quadrant of the coordinate plane, which makes things a bit simpler visually. Each constraint represents a line, and the inequalities tell us which side of the line our feasible region lies on. For instance, x + y ≤ 5 means we're looking at the area below or on the line x + y = 5.

Why Linear Programming Matters

Linear programming isn't just some abstract math concept; it's a powerful tool used in countless real-world applications. Businesses use it to optimize production schedules, minimize costs, and maximize profits. Airlines use it to plan flight routes and crew schedules. Logistics companies use it to optimize delivery routes. Even things like portfolio optimization in finance rely on linear programming principles. By understanding the core ideas, we can apply them to a wide range of problems. This is why mastering these techniques is so valuable, guys. It opens doors to solving practical problems in almost any field.

Step 1: Graphing the Constraints

The first step in solving this linear programming problem is to visualize the feasible region. We do this by graphing the constraints. Each constraint represents a line, and the inequalities tell us which side of the line is included in the feasible region. Remember, we're looking for the area where all the constraints are satisfied simultaneously.

Converting Inequalities to Equations

To graph the lines, we first convert the inequalities into equations:

x + y ≤ 5 becomes x + y = 5
x - y ≥ -1 becomes x - y = -1
x ≥ 0 is the y-axis
y ≥ 0 is the x-axis

Plotting the Lines

Now, let's plot these lines on a coordinate plane. To plot a line, we need two points. We can find these by choosing convenient values for x and solving for y, or vice versa.

For x + y = 5:
- If x = 0, then y = 5. So, we have the point (0, 5).
- If y = 0, then x = 5. So, we have the point (5, 0).
- Connect these points to draw the line.
For x - y = -1:
- If x = 0, then -y = -1, so y = 1. We have the point (0, 1).
- If y = 0, then x = -1. We have the point (-1, 0). However, since we also have the constraints x ≥ 0 and y ≥ 0, we're only interested in the portion of the line in the first quadrant or on the axes.
- Connect these points to draw the line.
x = 0 is simply the y-axis.
y = 0 is simply the x-axis.

Determining the Feasible Region

Now comes the crucial step: figuring out which side of each line is included in the feasible region. We can do this by testing a point (like (0, 0)) in each inequality. If the inequality holds true, then the feasible region is on the same side of the line as our test point. If it's false, the feasible region is on the opposite side.

For x + y ≤ 5: Plugging in (0, 0), we get 0 + 0 ≤ 5, which is true. So, the feasible region is below the line x + y = 5.
For x - y ≥ -1: Plugging in (0, 0), we get 0 - 0 ≥ -1, which is true. So, the feasible region is above the line x - y = -1.
x ≥ 0 means we're to the right of the y-axis.
y ≥ 0 means we're above the x-axis.

The feasible region is the area where all these conditions are met – it's the polygon formed by the intersection of all the shaded regions. It's super important to shade this region clearly, guys, as it's the visual representation of all possible solutions to our problem.

Step 2: Identifying the Vertices

The next step is to identify the vertices (corner points) of the feasible region. These are the points where the constraint lines intersect. Remember, the maximum or minimum value of our objective function will always occur at one of these vertices. So, finding them accurately is absolutely crucial. If you get the vertices wrong, the whole solution goes sideways!

Reading Vertices from the Graph

Sometimes, you can simply read the coordinates of the vertices directly from your graph. However, to be absolutely sure of their accuracy, especially when the lines intersect at non-integer points, we'll often need to solve the system of equations formed by the intersecting lines.

Finding Vertices by Solving Systems of Equations

Let's identify the vertices of our feasible region:

Intersection of x + y = 5 and x - y = -1:

We can solve this system of equations using elimination or substitution. Let's use elimination:

Add the two equations:

(x + y) + (x - y) = 5 + (-1)

2x = 4

x = 2

Substitute x = 2 into x + y = 5:

2 + y = 5

y = 3

So, the vertex is (2, 3).
Intersection of x + y = 5 and y = 0:

Substitute y = 0 into x + y = 5:

x + 0 = 5

x = 5

So, the vertex is (5, 0).
Intersection of x - y = -1 and x = 0:

Substitute x = 0 into x - y = -1:

0 - y = -1

y = 1

So, the vertex is (0, 1).
Intersection of x = 0 and y = 0:

This is simply the origin, (0, 0).

So, our vertices are (0, 0), (0, 1), (2, 3), and (5, 0). Double-check your calculations, guys, to make sure you've got these right! A small error here can throw off your final answer.

Step 3: Evaluating the Objective Function at Each Vertex

Now that we've identified the vertices of our feasible region, the next step is to evaluate the objective function, P = 24x + 30y, at each of these points. This will tell us the value of P at each corner of our feasible region. Remember, the maximum (or minimum) value of P will always occur at one of these vertices. It's like we're testing each possible "best" solution to see which one really wins.

Plugging in the Vertices

Let's plug each vertex into the objective function:

At (0, 0):

P = 24(0) + 30(0) = 0
At (0, 1):

P = 24(0) + 30(1) = 30
At (2, 3):

P = 24(2) + 30(3) = 48 + 90 = 138
At (5, 0):

P = 24(5) + 30(0) = 120

Interpreting the Results

We've now calculated the value of P at each vertex. We have P = 0 at (0, 0), P = 30 at (0, 1), P = 138 at (2, 3), and P = 120 at (5, 0). To find the maximum value of P, we simply look for the largest value among these results. In this case, the maximum value is 138.

Step 4: Determining the Maximum Value and Solution

After evaluating the objective function at each vertex, we can now determine the maximum value of P and the corresponding values of x and y that achieve this maximum. This is the grand finale, guys – the moment we answer the original question!

Identifying the Maximum Value

From our calculations in the previous step, we found the following values for P:

At (0, 0): P = 0
At (0, 1): P = 30
At (2, 3): P = 138
At (5, 0): P = 120

The largest of these values is 138. Therefore, the maximum value of P is 138.

Stating the Solution

Now, let's state our solution clearly and completely. The maximum value of P = 24x + 30y, subject to the given constraints, is 138. This maximum value occurs when x = 2 and y = 3. It's super important to state both the maximum value and the values of x and y that give us that maximum. We're not just interested in what the maximum is, but where it happens.

Double-Checking the Solution

It's always a good idea to double-check our solution, guys, to make sure we haven't made any mistakes. We can do this by plugging x = 2 and y = 3 back into our constraints to ensure they're satisfied:

x + y ≤ 5: 2 + 3 ≤ 5 (True)
x - y ≥ -1: 2 - 3 ≥ -1 (-1 ≥ -1) (True)
x ≥ 0: 2 ≥ 0 (True)
y ≥ 0: 3 ≥ 0 (True)

Since all the constraints are satisfied, we can be confident that our solution is correct. We've successfully found the maximum value of P and the values of x and y that achieve it!

Conclusion: Mastering Linear Programming

So, there you have it, guys! We've successfully tackled a linear programming problem, finding the maximum value of P = 24x + 30y subject to a set of constraints. We walked through the process step-by-step, from graphing the constraints and identifying the feasible region to finding the vertices and evaluating the objective function. Remember, linear programming is a powerful tool with applications in many different fields, so mastering these techniques is well worth the effort. Keep practicing, and you'll become a linear programming pro in no time! This systematic approach ensures that we not only find the answer but also understand the why behind it. This understanding is what allows us to apply these principles to more complex and varied problems in the future. Keep up the great work, and remember to break down challenging problems into smaller, manageable steps. You've got this!