Projects # 1,3,5 Example

Essays on Projects # 1,3,5 Speech or Presentation Insert and box number here) MA 120 BID: 294 18 February Project 5: Linear Programming Applications An automobile manufacturer makes cars and trucks in a factory that is divided into two shops. Shop 1, which performs basic assembly, must work five man-days on each truck, but only two man-days on each car. Shop 2, which performs finishing operations, must work three man-days for each car or truck it produces. Because of men and machine limitations, Shop 1 has 180 man-days per week available, while Shop 2 has 135 man-days per week. If the manufacturer makes a profit of $300 on each truck and $200 on each car, how many of each should be produced to maximize profit?The variables when solving this particular equation include the man-days, and the machine limitations. The constraints presented in this particular problem are the man-days available per week, which vary depending on shop and task. The objective function of this particular equation is to determine the best way to maximize profi ts based upon vehicles produced. Solution: Let x be the number of trucks and y the number of cars to be produced on a weekly basis. 5x + 2y = 1803x + 3y = 135A= ( 5 2 3 3), B = (180 135), C = 300,200Maximum problem: The vector x must be determined so that the weekly profit, as detailed by quantity cx, is a maximum which is subject to the inequality constraints Ax = b and x = 0. The inequality constraints work to insure that the weekly number of available man-days is not exceeded, and that there are no non-negative quantities of automobiles or trucks produced. The graph of the convex set of possible x vectors is presented above. The extreme points of the convex set C are: T (0 0). T2 = (36 0), T3 = (0 45) and T4 (30 15)/To solve the equation the function cx= 300x +200y must be tested at each of these points. The values taken are 0, 10800, 9000, and 12000.The maximum weekly profit is $12,000 and is achieved by producing 30 trucks and 15 cars per week.