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Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Maximize \(\quad p=x-2 y\) subject to \(\begin{aligned} x+2 y & \leq 8 \\ x-6 y & \leq 0 \\ 3 x-2 y & \geq 0 \\ x \geq 0, y & \geq 0 . \end{aligned}\)

Short Answer

Expert verified
The optimal solution to the LP problem is \((8, 0)\) and the maximum value of the objective function \(p = x - 2y\) is 8.

Step by step solution

01

Identify the constraints in canonical form

The problem already gives us the constraints in canonical form. We will rewrite them here: 1. \(x + 2y \leq 8\) 2. \(x - 6y \leq 0\) 3. \(3x - 2y \geq 0\) 4. \(x \geq 0\) 5. \(y \geq 0\)
02

Plot the feasible region on the coordinate plane

To plot the feasible region, we first identify the inequalities and equalities representing the half-planes for each constraint. Then, we sketch the half-planes on the coordinate plane, shading the areas that satisfy all the given constraints simultaneously, forming the feasible region. 1. For \(x + 2y \leq 8\), the equality line is \(x + 2y = 8\). 2. For \(x - 6y \leq 0\), the equality line is \(x - 6y = 0\). 3. For \(3x - 2y \geq 0\), the equality line is \(3x - 2y = 0\). 4. For \(x \geq 0\), the equality line is \(x = 0\). 5. For \(y \geq 0\), the equality line is \(y = 0\).
03

Determine the vertices of the feasible region

We now find the intersections of the lines that form the feasible region to identify the vertices: 1. Intersection of \(x + 2y = 8\) and \(x - 6y = 0\) is \((3, 2.5)\). 2. Intersection of \(x + 2y = 8\) and \(3x - 2y = 0\) is \((2, 3)\). 3. Intersection of \(x - 6y = 0\) and \(3x - 2y = 0\) is not possible because the two lines are parallel. 4. Intersection of \(x + 2y = 8\) and \(x = 0\) is \((0, 4)\). 5. Intersection of \(x + 2y = 8\) and \(y = 0\) is \((8, 0)\). Thus, the vertices of the feasible region are: \((3, 2.5)\), \((2, 3)\), \((0, 4)\), and \((8, 0)\).
04

Evaluate the objective function at each vertex

We will evaluate the objective function \(p = x - 2y\) at each vertex: 1. At \((3, 2.5)\), \(p = 3 - 2(2.5) = -2\). 2. At \((2, 3)\), \(p = 2 - 2(3) = -4\). 3. At \((0, 4)\), \(p = 0 - 2(4) = -8\). 4. At \((8, 0)\), \(p = 8 - 2(0) = 8\).
05

Identify the maximum value of the objective function and the optimal solution

From the evaluated objective function values: 1. At \((3, 2.5)\), \(p = -2\). 2. At \((2, 3)\), \(p = -4\). 3. At \((0, 4)\), \(p = -8\). 4. At \((8, 0)\), \(p = 8\). We see that the maximum value of the objective function is 8, which occurs at the point \((8, 0)\). Thus, the optimal solution is \((8, 0)\) and the maximum value of the objective function \(p = x - 2y\) is 8.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Feasible Region
The feasible region in an LP (Linear Programming) problem is the set of all possible points that satisfy all the constraints of the problem. It's like finding a safe zone on a map that obeys certain rules.

For our exercise, the rules are the inequalities given, such as: \(x+2y \text{≤} 8\), \(x-6y \text{≤} 0\), \(3x-2y \text{≥} 0\), and that both \(x\) and \(y\) are not negative; you can't have a negative number of items or distances in typical situations, right? This area is crucial because any solution that lies outside of this 'safe zone' isn't valid for our problem.

To visualize it, imagine drawing a boundary for every rule on a graph. The space where all these areas overlap is our feasible region. For a better understanding, think of each equation as a fence in a field, and the feasible region is the space within all fences. In step 2 of the solution, we sketch these 'fences' on a plane, and the common area shaded forms the feasible region. Next, by calculating the intersection of these lines— our fence corners— we can find the vertices of this special zone, as shown in step 3.
Objective Function
Think of the objective function as your ultimate goal or target in an LP problem. It's what you want to achieve, minimize or maximize, given the conditions you're working with.

In our exercise, the objective function is to maximize \(p = x - 2y\). It symbolizes the measure of success or outcome that we are aiming for, be it profit, cost-saving, distance, or anything else pertinent to our problem. Evaluating the objective function means plugging our 'coordinates' into the equation to see how 'successful' each point is.

In step 4, we find out how good each vertex of our feasible region is by substituting its values into our objective function. It's like checking which point in the 'safe zone' gives us the highest score or the most profit. We do this knowing that the optimal solution to an LP problem can often be found at one of these vertices, thanks to the properties of linear functions and constraints.
Optimization
Optimization in the context of LP problems is the process of finding the best possible answer— the highest or lowest value of the objective function within the feasible region. It's like playing a video game and trying to get the highest score or finding the best deal when shopping on a budget.

During optimization, we compare the scores we got for each point in our feasible region. This comparison lets us know which point, if any, gives us our 'high score', which in this case is maximizing \(p\). In step 5, we find out that among all the vertices we tested, the point \((8, 0)\) gives us the highest value of \(p\) which is 8.

This makes \((8, 0)\) our optimal solution, and achieving a score of 8 is our best outcome. With this value, we've optimized our objective function by finding the maximum point within our defined conditions. Optimization helps to ensure that resources are used efficiently and objectives are met in the most effective way, reflecting the ideal scenario based on the model's constraints.

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