Maximize Z = 3x + 4y: A Step-by-Step Guide

Hey guys! Today, we're diving into a super cool problem from the world of linear programming. We're going to figure out how to maximize the value of the expression z = 3x + 4y given a set of constraints. Think of it like finding the sweet spot within a specific area – the point that gives us the highest possible value for our target expression. This is a fundamental concept with tons of real-world applications, from optimizing business operations to resource allocation. So, buckle up, and let's get started!

Understanding the Problem

First things first, let's break down exactly what we're trying to do. We want to find the maximum value of z = 3x + 4y. This expression is called our objective function. The values of x and y can't be just anything; they have to satisfy certain conditions, known as constraints. These constraints define the feasible region, which is the set of all possible (x, y) pairs that meet all the conditions. Our constraints are:

• x ≥ 0 (x is non-negative)
• y ≥ 0 (y is non-negative)
• y - x - 1 ≤ 0 (or y ≤ x + 1)
• y - x + 1 ≥ 0 (or y ≥ x - 1)
• x + y ≤ 5

These constraints define a specific area on a graph, and we need to find the point within that area that makes z as big as possible. Think of it as navigating a maze where only certain paths are allowed, and you want to find the path that leads to the biggest treasure!

Visualizing the Constraints: Graphing the Feasible Region

To truly understand what's going on, let's visualize these constraints. We'll graph them on the x-y plane. Each inequality represents a region, and the feasible region is where all these regions overlap. Trust me, sketching this out will make the whole problem much clearer!

1. x ≥ 0 and y ≥ 0: These are the simplest constraints. They tell us that we're only interested in the first quadrant (where both x and y are positive or zero).
2. y ≤ x + 1: This is a line with a slope of 1 and a y-intercept of 1. The region we're interested in is below this line. To plot it, first draw the line y = x + 1. Pick two points, say (0, 1) and (1, 2), and connect them. Then, shade the area below the line because we want y to be less than or equal to x + 1.
3. y ≥ x - 1: This is another line with a slope of 1, but this time the y-intercept is -1. The region we want is above this line. Draw the line y = x - 1 (for example, using the points (0, -1) and (1, 0)). Shade the area above this line.
4. x + y ≤ 5: This line has a slope of -1 and intercepts the axes at (5, 0) and (0, 5). We want the region below this line. Draw the line x + y = 5 (points (5, 0) and (0, 5) can be used) and shade below it.

When you graph all these inequalities, you'll see that they create a closed shape – a pentagon in this case. This pentagon is our feasible region. Any point inside or on the boundary of this pentagon satisfies all our constraints. The next step is figuring out which of these points gives us the maximum value of z.

The Corner Point Theorem: Our Secret Weapon

Now comes the magic part! There's a fundamental principle in linear programming called the Corner Point Theorem. This theorem states that the maximum (and minimum) value of a linear objective function, like our z = 3x + 4y, will always occur at one of the corner points (also called vertices) of the feasible region. This is huge because it drastically reduces our work. Instead of checking every single point within the pentagon, we only need to check the coordinates of its corners!

In our case, the feasible region is described as a convex (closed) pentagon with vertices O(0,0), P1(0,1), P2(2,3), P3(3,2), and P4(1,0). These are the only points we need to evaluate.

Finding the Vertices: Where the Lines Meet

To apply the Corner Point Theorem, we need to identify the vertices (corner points) of our feasible region. These are the points where the constraint lines intersect. By solving the equations of the intersecting lines, we can find the coordinates of these vertices. This is essential for finding the optimal solution in linear programming problems.

Let's go through how to find these points:

1. The vertex O(0, 0): This is the easiest one! It's the origin, where the lines x = 0 and y = 0 intersect.

2. The vertex P1(0, 1): This is the intersection of the lines x = 0 and y = x + 1. Substituting x = 0 into y = x + 1, we get y = 1.

3. The vertex P2(2, 3): This is the intersection of the lines y = x + 1 and x + y = 5. We can use substitution here. From y = x + 1, substitute this into the second equation: x + (x + 1) = 5. Simplify to get 2x + 1 = 5, then 2x = 4, so x = 2. Now, plug x = 2 back into y = x + 1 to get y = 3.

4. The vertex P3(3, 2): This is the intersection of the lines y = x - 1 and x + y = 5. Again, use substitution. From y = x - 1, substitute into the second equation: x + (x - 1) = 5. Simplify to get 2x - 1 = 5, then 2x = 6, so x = 3. Now, plug x = 3 back into y = x - 1 to get y = 2.

5. The vertex P4(1, 0): This is the intersection of the lines y = 0 and y = x - 1. Substituting y = 0 into y = x - 1, we get 0 = x - 1, so x = 1.

So, we've successfully identified all five vertices of our feasible region: O(0, 0), P1(0, 1), P2(2, 3), P3(3, 2), and P4(1, 0). These are our contenders for the point that maximizes z!

Evaluating the Objective Function: Finding the Winner

We've found our vertices; now it's time for the final showdown! We need to plug the coordinates of each vertex into our objective function, z = 3x + 4y, and see which one gives us the largest value. This is the crucial step in determining the optimal solution in our linear programming problem.

Let's do the calculations:

• At O(0, 0): z = 3(0) + 4(0) = 0
• At P1(0, 1): z = 3(0) + 4(1) = 4
• At P2(2, 3): z = 3(2) + 4(3) = 6 + 12 = 18
• At P3(3, 2): z = 3(3) + 4(2) = 9 + 8 = 17
• At P4(1, 0): z = 3(1) + 4(0) = 3

By comparing the values of z at each vertex, it's clear that the maximum value of z is 18, which occurs at the point P2(2, 3). This means that the optimal solution to our problem is x = 2 and y = 3, giving us a maximum value of 18 for the objective function z = 3x + 4y.

Conclusion: The Maximum Value Unveiled

So, there you have it! We've successfully maximized z = 3x + 4y subject to the given constraints. The maximum value is 18, and it occurs at the point (2, 3). We used a combination of graphing, the Corner Point Theorem, and simple algebra to solve this problem. Linear programming can seem intimidating at first, but by breaking it down step by step, it becomes much more manageable. The key is to visualize the constraints, identify the feasible region, find the vertices, and then evaluate the objective function at those vertices.

This method is incredibly useful in various real-world scenarios, such as optimizing production, resource allocation, and even investment strategies. So, keep practicing, and you'll become a linear programming pro in no time! Remember, understanding these concepts not only helps you solve math problems but also provides valuable tools for decision-making in various aspects of life. Keep exploring and applying these techniques, and you'll be amazed at the power of mathematical optimization!