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Linear Programming

  1. Oct 31, 2008 #1
    A furniture manufacturer makes chairs and sofas. Each chair can be sold for a profit of £15 and each sofa for a profit of £5. It takes 4 hours to make the chair and 5 hours to make the sofa. The manufacturer has enough workers to provide 200 hours per week producing the furniture. Customer demand requires that at least 7 times as many chairs as sofas are made. Chairs take up 1 m^3 of storage space and sofas take up 3 m^3. There is a total of 90m^3 of storage space available in the factory per week.

    Q1. What is the weekly profit of the manufacturer, assuming that the demand for furniture means that all item will be sold?

    Q2. What are the constraints for the problem??

    Q3. Solve the linear programming.

    Anyone good with linear programming???
    Last edited: Oct 31, 2008
  2. jcsd
  3. Oct 31, 2008 #2


    Staff: Mentor

    What have you done already? Have you looked at a worked example in your book?
  4. Nov 1, 2008 #3
    Yea I have found the profit and the constraints, just don't know how to draw the graph showing the optimal point, iso-profit and optimal profit.
  5. Nov 1, 2008 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Then show us what you have. The graphs of the constraints are just straight lines. That's always true in Linear Programming.
  6. Nov 1, 2008 #5
    P= 15x + 5y subject to

    4x + 5y > 200
    1x + 3y > 90
    x > 0
    y > 0

    That's my constraints
  7. Nov 1, 2008 #6


    Staff: Mentor

    Presumably x = number of chairs and y = number of sofas.

    Your first two inequalities go the wrong way: the total time has to be <= 200, and the total volume has to be <= 90.

    The last two inequalities go the right way but need to include 0.

    There is one inequality that you're missing- the one about customer demand. If you graph the four inequalities you have plus the one you're missing you should be able to find the corner point that maximizes profit.
  8. Nov 1, 2008 #7
    How do I do the customer demand constraint???
  9. Nov 1, 2008 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    Your problem said "Customer demand requires that at least 7 times as many chairs as sofas are made." You are using x to represent the number of chairs made and y to represent the number of sofas made (according to Mark44- you didn't tell us that, yourself!) so [itex]x\ge 7y[/itex].

    Now, where do all the lines bounding those inequalities intersect? What are the vertices of the "feasible region"?
  10. Nov 2, 2008 #9
    I am not very good with graph drawing, so finding the linear programming problem graphically and finding the feasible region, optimal point would be difficult for me. Are there any easy way of finding these after obtaining the constraints??
  11. Nov 2, 2008 #10


    Staff: Mentor

    Difficult, maybe, but not impossible. Each and every one of your constraints is a linear inequality, so if you can draw the graph of a line, you're almost done. The difference between these inequalities and the associated linear equations is that each inequality determines a half-plane.

    For each inequality draw the line that is the boundary. After that decide which side of the line satisfies the inequality and shade that side of the line.

    For example, one of your constraints is x + 3y <= 90.

    Can you draw the line whose equation is x + 3y = 90? Its slope is -1/3 and its y-intercept is 30 (meaning the point (0, 30) is on the graph of the line).

    After drawing the line pick any point that is not on the line and see whether it satisfies the inequality x + 3y < 90. If it does that entire side of the line and the line is your solution set for that inequality. If the point you picked doesn't satisfy the inequality, the other side of the line (and the line) is the solution set for the inequality.

    You've said you're not very good at drawing graphs. It looks like you're in some sort of math class, so here's your opportunity to get better at it. Just because you're not good at something doesn't mean that will always have to be true.
  12. Nov 2, 2008 #11
    Thanks a lot, will do that now
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