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Math Help - Linear programming

  1. #1
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    Linear programming

    Hi.
    I'm trying to solve a linear programming task (using the graphing method) for profit maximization.
    Say, there's a small business producing peanut and raisin mixtures called PR and RP.
    The mixture PR contains 2/3 kg raisins and 1/3 kg peanuts, whereas mixture RP contains 1/2 kg raisins and 1/2 kg peanuts. The produce these mixtures, the businessman can buy 90 kg raisins and 60 kg of peanuts, with 1 kg of peanuts costing 0.6 and 1 kg of raisins costing 1.5. Packages labelled PR are sold at a price of 2.9 /kg and packages labelled RP 2.55 /kg.
    The owner assumes that no more than 150 mixture packages will be sold.
    How many different mixture packages should the small business produce to maximise profit?

    I figure the variables should be: x1 - number of PR mixture packages produced, x2 - number of RP mixture packages produced.
    The constraints, I think, would be: 1/3x1 + 1/2x2 <= 60 (maximum amount of peanuts)
    2/3x1 + 1/2x2 <= 90 (maximum amount of raisins)
    x1 + x2 <= 150 (maximum amount of mix packages)
    Anything else?
    The optimization equation is what I can't quite figure out, though. I figure I could just write it F = 2.9x1 + 2.55x2 --> max. But this formula would calculate revenue maximisation, not profit. So any way to make the formula take costs into account as well?

    I did draw the graph (the feasibility region) and point of maximization does in fact show (90;60), which should be the correct answer. Point is, how to use the optimization equation to calculate the comparative values of all the corner points of the feasibility region?
    Last edited by JacobE; February 8th 2013 at 09:48 AM.
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  2. #2
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    Re: Linear programming

    Quote Originally Posted by JacobE View Post
    Hi.
    I'm trying to solve a linear programming task (using the graphing method) for profit maximization.
    Say, there's a small business producing peanut and raisin mixtures called PR and RP.
    The mixture PR contains 2/3 kg raisins and 1/3 kg peanuts, whereas mixture RP contains 1/2 kg raisins and 1/2 kg peanuts. The produce these mixtures, the businessman can buy 90 kg raisins and 60 kg of peanuts, with 1 kg of peanuts costing 0.6 and 1 kg of raisins costing 1.5. Packages labelled PR are sold at a price of 2.9 /kg and packages labelled RP 2.55 /kg.
    The owner assumes that no more than 150 mixture packages will be sold.
    How many different mixture packages should the small business produce to maximise profit?

    I figure the variables should be: x1 - number of PR mixture packages produced, x2 - number of PR mixture packages produced.
    Typo- that second "PR" should be "RP".

    The constraints, I think, would be: 1/3x1 + 1/2x2 <= 60 (maximum amount of peanuts)
    2/3x1 + 1/2x2 <= 90 (maximum amount of raisins)
    x1 + x2 <= 150 (maximum amount of mix packages)
    Anything else?
    Only that x1+ x2>0 0.

    The optimization equation is what I can't quite figure out, though. I figure I could just write it F = 2.9x1 + 2.55x2 --> max. But this formula would calculate revenue maximisation, not profit. So any way to make the formula take costs into account as well?
    Yes, subtract off the costs. If you use (1/3)x1 kg of peanuts in RP and (1/2)x2 kg of peanuts in Pr, and each kilogram of peanuts cost 0.6 then you need to subtract .6((1/3)x_1+ (1/2)x_2) from the profit. Do the same for the raisins and packages.

    I did draw the graph (the feasibility region) and point of maximization does in fact show (90;60), which should be the correct answer. Point is, how to use the optimization equation to calculate the comparative values of all the corner points of the feasibility region?
    Great. But "point of maximization" of what. The point you want should be the vertex where the profit is largest and you are saying you do not know how to find that.
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  3. #3
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    Re: Linear programming

    Thanks for the input.
    So, this is what I've got so far. As I've understood it, producing a package of RP costs 1.5 * 2/3 + 0.6 * 1/3 = 1.2 and producing a package of PR costs 1.5 * 1/2 + 0.6 * 1/2 = 1.05.
    Hence the optimization equation for profit would be F = (2.9 - 1.2)x1 + (2.55 - 1.05)x2 = 1.7x1 + 1.5x2. Could this be correct?

    And also, the vertices, I've found them to be (0;120), (90;60) and (135;0), making profits 180, 243 and 229.5 respectively. So the vertex where profit is maximised should be (90;60). I was originally told to be getting a value of 270 at (90:60) but I'm quite confident I'm on the right track at the moment. Hint maybe?
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  4. #4
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    Re: Linear programming

    The reason I asked is because I wasn't quite sure whether I calculated the costs per unit correctly. Anyone care to confirm or refute my progress?
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