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Math Help - Linear Program help

  1. #1
    Rob
    Rob is offline
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    Linear Program help

    Hello, I'm stumped on this problem. Any help would be greatly appreciated Also a good explanation would be nice, considering I suck at math. Thanks!

    The publicity director for Mercy Hospital is planning to
    bolster the hospital’s image by running a TV ad and a radio ad. Due
    to budgetary and other constraints, the number of times she can run
    the TV ad (denoted by x) and the number of times she can run the
    radio ad (y), is constrained to lie in the region shown below (the
    region of feasible solutions is given the portion enclosed by the line
    segments). The formula


    A = 90000x+ 40000y

    gives the total number of people reached by the ads. Find the maximum number

    of people that can be reached running these ads. What
    is the best mix for the number of television and number of radio ads
    to run?




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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Rob View Post
    Hello, I'm stumped on this problem. Any help would be greatly appreciated Also a good explanation would be nice, considering I suck at math. Thanks!

    The publicity director for Mercy Hospital is planning to
    bolster the hospitalís image by running a TV ad and a radio ad. Due
    to budgetary and other constraints, the number of times she can run
    the TV ad (denoted by x) and the number of times she can run the
    radio ad (y), is constrained to lie in the region shown below (the
    region of feasible solutions is given the portion enclosed by the line
    segments). The formula


    A = 90000x+ 40000y

    gives the total number of people reached by the ads. Find the maximum number

    of people that can be reached running these ads. What

    is the best mix for the number of television and number of radio ads
    to run?






    With a linear objective to maximise/minimise we know that the maximum/
    minimum lies on the boundary of the feasible region. Assuming the boundary
    is a closed polygon we know in addition that the maximum/minimum is
    realised at a vertex (in the degenerate case an entire edge will be optimal
    but that still give two vertices where the objective is at its
    maximum/minimum.

    So if we evaluate the objective at each of the vertices of the feasible
    region one where it takes its greatest value is an optimal point.

    RonL
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