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Math Help - Optimization

  1. #1
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    Optimization

    Hello, at first had this problem:

    A mountain lake in national park is stocked each spring with two species of fish, S1 and S2. The average weight of the fish stocked is 4 pounds for S1 and 2 pounds for S2. Two foods, F1 and F2, are available in the lake. The average requeriment of a fish of species S1 is 1 unit of F1 and 3 units of F2 each day. The corresponding requeriment of S2 is 2 units of F1 and 1 unit of F2.

    a). Suppose that 1000 units of F1 and 1800 units of F2 are available daily in example previous. How should the lake be stocked to maximize the weight of fish supported by the lake?

    Well, I have already found the solution, graphical method using the search

    Maximize 4x_1 + 2x_2

    subject to restrictions

      x_1 + 2x_2 \leq{1000}

      3x_1 + x_2 \leq{1800}


    And the solution is   x_1 = 520, x_2 = 240


    The question I have is to resolve this part:

    In problem previous, how should the lake stocked to maximize the total number of fish supported by the lake?


    In advance thanks for helping answer this question.
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  2. #2
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    THe second part can be answered graphically as well, but here is an algebraic solution.
    We have,
    x_1+2x_2\leq1000 and 3x_1+x_2\leq1800.
    Add twice the first inequality to the second and you have
    5x_1+5x_2\leq3800.
    Divide by 5 and you get
    x_1+x_2\leq760.
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  3. #3
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    But keep in mind that this problem comes third restriction??
    <br />
4x_1 + 2x_2 = 2560.
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  4. #4
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    That isn't a problem is it ?
    x_{1}=520, \quad x_{2}=240, is at the limit of x_{1}+x_{2}\leq760.
    What this is saying is that the two alternative requirements, (1) maximize the weight and (2) maximize the number, lead to the same result.
    Am I missing something ?
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  5. #5
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    Hello,

    I think it should be as you say

    but I was posing like this: Since the first part of the exercise was found the maximum weight supported by the lake

    4(520) + 2(240) = 2560

    so now the problem is

    maximize  P = x_1 + x_2

    subject to restrictions

    4x_1 + 2x_2 = 2560

    x_1+2x_2\leq1000

    3x_1+x_2\leq1800.


    this is wrong, or gives the same solution?

    Thank you very much ...
    Last edited by Dogod11; November 11th 2009 at 05:28 AM.
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  6. #6
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    The result I get is P = 760,

    x = 520, y = 240


    well, anyway it does not matter jeje

    Greetings
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  7. #7
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    Quote Originally Posted by Dogod11 View Post
    Hello, at first had this problem:

    A mountain lake in national park is stocked each spring with two species of fish, S1 and S2. The average weight of the fish stocked is 4 pounds for S1 and 2 pounds for S2. Two foods, F1 and F2, are available in the lake. The average requeriment of a fish of species S1 is 1 unit of F1 and 3 units of F2 each day. The corresponding requeriment of S2 is 2 units of F1 and 1 unit of F2.

    a). Suppose that 1000 units of F1 and 1800 units of F2 are available daily in example previous. How should the lake be stocked to maximize the weight of fish supported by the lake?

    Well, I have already found the solution, graphical method using the search

    Maximize 4x_1 + 2x_2

    subject to restrictions

     x_1 + 2x_2 \leq{1000}

     3x_1 + x_2 \leq{1800}


    And the solution is  x_1 = 520, x_2 = 240


    The question I have is to resolve this part:

    In problem previous, how should the lake stocked to maximize the total number of fish supported by the lake?.
    You now have to maximise x_1+x_2 subject to the same constraints:

     x_1 + 2x_2 \leq{1000}

     3x_1 + x_2 \leq{1800}


    CB
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Dogod11 View Post
    Hello,

    I think it should be as you say

    but I was posing like this: Since the first part of the exercise was found the maximum weight supported by the lake

    4(520) + 2(240) = 2560

    so now the problem is

    maximize P = x_1 + x_2
    Yes

    subject to restrictions

    4x_1 + 2x_2 = 2560
    No, you cannot generaly get the maximum mass of fish with the maximum number.


    x_1+2x_2\leq1000

    3x_1+x_2\leq1800


    this is wrong, or gives the same solution?

    Thank you very much ...
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  9. #9
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    anyway gives the same result: x = 520, y = 240.


    Thank you very much....
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