Homework Help Forum: Linear Programming

Posted by Deanna on Friday, October 16, 2009 at 8:30pm.

OKay My question is: Two factories manufacture 3 different grades per paper. the company that owns the factories has contracts to supply at least 16 tons of low grade, 5 tons of medium grade, and at least 20 tons of high grade paper. It cost $1000 per day to operate the first factory and 2000 per day to operate the second. Factory 1 produces 8 tons of low grade, 1 ton of medium grade, and 2 tons of high grade paper in day's operation. Factory 2 produces 2 tons of low grade, 1 ton of medium grade and 7 tons of high grade per day. how many days should each factory be in operation in order to fill the most economically? Also i have to assign variables to x and y and state a goal plus represent the goal with an equation then i have to write inequalities that represent the restrictions of this problem in a chart. Then i have to grade the system of inequalities, shade the solution region, and use corner points to analyze data and find the solution. Finally I write my solution in a complete sentence. Help?

• Linear Programming - MathMate, Friday, October 16, 2009 at 8:41pm

  So the instructions seem to be clear. Do you have a problem, or where is the problem if there is any?

  
  

• Linear Programming - Deanna, Friday, October 16, 2009 at 8:59pm

  It's just I'm stuck at picking out what i need and such.. Not very good with word problems..

  
  

• Linear Programming - MathMate, Friday, October 16, 2009 at 11:07pm

  You can take the question and split it up into constraints and costs. Requirements: "supply at least 16 tons of low grade, 5 tons of medium grade, and at least 20 tons of high grade paper." So minimum quantities = (16,5,20) Supplied by Factory 1, F1 = (8,1,2)/$1000 Supplied by F2 = (2,1,7)/$2000 So the constraints are: 8F1+2F2 ≥ 16 F1+F2 ≥ 5 2F1+7F2 ≥ 20 F1 ≥ 0 F2 ≥ 0 Cost=1000F1+2000F2 Your graph will have F1 in the x-axis, F2 in the y-axis. Plot the constraint equations, and shade the solution (feasible) region. Find the optimal solution.

  
  

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