linear optimization/ Matlab coding Part

(a) An electronics company must deliver exactly 2400 units within the next 3 weeks. The client will pay $30 for each unit delivered by the end of the first week; $25 for each unit delivered by the end of the second week, and $20 for each unit delivered by the end of the third week. Each worker can only assemble 40 units per week. The present labour force of the company is 15 workers. Hence, the company must hire and train temporary help. Any of the experienced workers can be taken off the assembly line to instruct at most two trainees; after one week of instruction, each of the trainees can either proceed to the assembly line or instruct new trainees. Some workers may become idle towards the end of the project; however, all workers must be kept on the payroll until the end of the project. The weekly wages of a worker, whether assembling, instructing, or being idle, are $600; the weekly wages of a trainee are $300. We assume that wages are the only costs associated with production. Here is one possible example of scheduling the production: 1 Week 1: 12 assemblers, 3 instructors, 6 trainees, revenue from assembled units = 12 40 30 = 14400, wages = 12 600 + 3 600 + 6 300 = 10800. Week 2: 13 assemblers, 8 instructors, 16 trainees, revenue from assembled units = 13 40 25 = 13000, wages = 13 600 + 8 600 + 16 300 = 17400. Week 3: 35 assemblers, 2 idle, revenue from assembled units = 35 40 20 = 28000, wages = 35 600 + 2 600 = 22200. Total profit = total revenue total wages = 55400 50400 = 5000 dollars. The company wants to schedule the production so as to maximize the total profit. Write down a Linear Programming formulation of this problem as Matlab code. Explain the meaning of your variables. Hint: It follows from the theory of Linear Programming that you do not need to explicitly assume that the variables in your formulation of this problem take only integer values. Part (b) Solve the problem in Matlab. Print an optimal schedule of production and the total profit obtained from this optimal solution