/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A manufacturer produces two mode... [FREE SOLUTION] | 91Ó°ÊÓ

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A manufacturer produces two models of elliptical cross-training exercise machines. The times for assembling, finishing, and packaging model \(\mathrm{X}\) are 3 hours, 3 hours, and 0.8 hour, respectively. The times for model Y are 4 hours, 2.5 hours, and 0.4 hour. The total times available for assembling, finishing, and packaging are 6000 hours, 4200 hours, and 950 hours, respectively. The profits per unit are \(300\) for model \(X\) and \(375\) for model Y. What is the optimal production level for each model? What is the optimal profit?

Short Answer

Expert verified
The exact values of \(x\) and \(y\) ( optimal production level for each model) depend on solving the system of inequalities given above and thus cannot be immediately provided. The optimal profit can be found by substitifying the optimal values of \(x\) and \(y\) into the objective function Z = 300x + 375y.

Step by step solution

01

Setting up the Inequalities

Let \(x\) and \(y\) represent the units of model X and model Y to be produced, respectively. The time spent on assembling, finishing, and packaging can not exceed the total time available for these processes. Therefore, these constraints can be represented by the following inequalities: 3x + 4y ≤ 6000 (assembling), 3x + 2.5y ≤ 4200 (finishing), 0.8x + 0.4y ≤ 950 (packaging). Also, \(x\) and \(y\) must be non-negative since we cannot produce negative units: \(x\) ≥ 0 and \(y\) ≥ 0.
02

Setting up the Objective Function

The profit for model X is $300 and for model Y is $375. The manufacturer’s goal is to maximize this profit. Therefore, the objective function is Z = 300x + 375y.
03

Finding the Optimal Solution

The optimal solution lies at the intersection of the constraint lines that gives the maximum value for the objective function. This system of inequalities needs to be solved graphically or by using a linear programming algorithm, such as the simplex method, to find the values of \(x\) and \(y\) that maximize Z.

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Key Concepts

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

Objective Function
In linear programming, the objective function represents the goal of the problem, which is often to maximize or minimize a particular quantity. For this problem, the objective function seeks to maximize profit. Profit is calculated based on the number of products manufactured. Here, we have two types of products: model X and model Y.

Each unit of model X contributes \(300 to the profit, while each model Y contributes \)375. To express the total profit, we set up the objective function as:
  • \[ Z = 300x + 375y \]
Where:
  • \(x\) is the number of units of model X produced,
  • \(y\) is the number of units of model Y produced.
This objective function helps in determining how many units of X and Y should be produced to achieve the maximum possible profit.
Constraints
Constraints in linear programming are the limitations or requirements that must be met within a given problem. They can include restrictions on resources such as time, cost, or materials.

In this exercise, constraints are imposed by the total available hours for each process category: assembling, finishing, and packaging.
  • Assembling: The inequality for available assembling time is \(3x + 4y \leq 6000\).
  • Finishing: The inequality for available finishing time is \(3x + 2.5y \leq 4200\).
  • Packaging: The inequality for available packaging time is \(0.8x + 0.4y \leq 950\).
Furthermore, the non-negativity constraints, \(x \geq 0\) and \(y \geq 0\), ensure that a negative number of units cannot be produced. These inequalities define the feasible region within which an optimal solution will be found.
Profit Maximization
Profit maximization is the main goal in many linear programming problems. It refers to finding the production levels that lead to the highest possible profit. For this, one must determine the number of units of each product that should be produced within the constraints.

The profit, represented by the objective function \( Z = 300x + 375y \), is maximized when the production is adjusted such that all applicable constraints are satisfied. The solution process involves ensuring all scheduled assembly, finishing, and packaging tasks do not exceed the available time.
  • The graphical method can be applied by plotting the constraints and finding the vertex of the feasible region that gives the highest value of \(Z\).
  • Alternatively, the simplex method can be used to efficiently explore the vertices of the feasible region where optimal production levels will be found.
Simplex Method
The simplex method is an algorithm used to solve linear programming problems. It's especially helpful for problems that involve many constraints and variables.

In the context of this exercise, the simplex method would iterate through possible solutions at the corners (vertices) of the feasible region defined by the constraints.
  • The method begins with an initial feasible solution, usually one of the corners of the feasible region.
  • It then moves along the edges of the region to adjacent vertices to find a better solution.
  • This process continues until the maximum value of the objective function, \(Z\), is reached.
The advantage of the simplex method is its efficiency in handling complex linear programming problems, ensuring you reach the optimal solution without having to test every single possible combination of variables.

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Most popular questions from this chapter

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