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Math Help - linear programming, quantitive math

  1. #1
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    linear programming, quantitive math

    I have a question like this that will be on a test coming up soon and I need to know how to approach this problem. Basically I just need to know hot to set the problem up, you know, the coefficents variables and constraints. Once I have the problem together I can solve it, but I just need to know how to configure it. Thanks



    An ad campaign for a new snack chip will be conducted in a limited geographical area and can use TV time, radio time, and newspaper ads. Information about each medium is shown below.

    Medium/ Cost Per Ad/ # Reached/ Exposure Quality
    TV /500 /10000/ 30
    Radio /200 /3000/ 40
    New /400 /5000/ 25

    If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is $10000, develop the model that will maximize the number reached and achieve an exposure quality of at least 1000.

    Objective function-To maximize the number reached
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  2. #2
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    Hello, jwells1999!

    An ad campaign for a new snack chip can use TV time, radio time, and newspaper ads.
    Information about each medium is shown below.

    \begin{array}{cccc}\text{Medium} & \text{Cost Per Ad} & \text{No. reached} & \text{Exp. Quality} \\ \hline<br />
\text{TV} & 500 & 10,000 & 	30 \\<br />
\text{Radio} & 200 & 3,000 & 40 \\<br />
\text{News} & 400 & 5,000 & 	25 \\ \hline<br />
\text{Total} & 10,000 & N & 1,000  \end{array}

    If the number of TV ads cannot exceed the number of radio ads by more than 4,
    and if the advertising budget is $10000, develop the model that will maximize
    the number reached and achieve an exposure quality of at least 1000.

    Objective function: To maximize the number reached

    Let x = number of TV ads.
    Let y = number of radio ads.
    Let z = number of newspaper ads.


    Then the chart looks like this:

    \begin{array}{cccc}\text{Medium} & \text{Cost Per Ad} & \text{No. reached} & \text{Exp. Quality} \\ \hline<br />
\text{TV (x)} & 500x & 10,000x & 	30x \\<br />
\text{Radio (y)} & 200y & 3,000y & 40y \\<br />
\text{News (z)} & 400z & 5,000z & 	25z \\ \hline<br />
\text{Total} & 10,000 & N & 1,000  \end{array}


    And we have these inequalities:

    . . . . . . \begin{array}{c} x\:\geq\:0 \\ y\:\geq\:0 \\ z\:\geq\:0 \\<br />
x\:\leq \:y+4 <br />
\end{array}

    \begin{array}{c} 500x+200y + 400z \:\leq \:10,000 \\ 30x + 40y + 50z \:\geq \:1,000<br />
\end{array}


    Objective function: . N \;=\;10,000x + 3,000y + 5,000z

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