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A small firm produces AM and AM/FM car radios. The AM radios take 15 hours to produce, and the AM/FM radios take 20 hours. The number of production hours is limited to 300 hours per week. The plant’s capacity is limited to a total of 18 radios per week, and existing orders require that at least 4 AM radios and at least 3 AM/FM radios be produced a week. Write a system of inequalities representing this situation. Then draw a graph of the feasible region given these conditions, in which x is the number of AM radios and y the number of AM/FM radios.

  • math - ,

    This seems to be linear programming night
    let number of AMs be x
    let number of FMs be y
    then constraints:
    15 x + 20 y </= 300 hour limit line
    x+y </= 18 number limit line
    x>/= 4
    y>/= 3

    graph
    the hour limit line hits the y axis at (0,15) and the x axis at (20,0)
    the number limit line hits the y axis at (0,18) and the x axis at (18,0)
    we have a vertical line at x = 4
    and a horizontal line at y =3
    now find intersections (vertices)
    (4,3)
    (4,48) intersection of x = 4 with hours limit line
    (12,6) intersection of hours line and numbers line
    (15,3) intersection of numbers line with horizontal line at y = 3

    Draw straight lines around the vertices and you have your region inside.

  • Linear programming - ,

    PS
    Label questions like this "linear programming" so people who do linear programming will help.

  • math - ,

    Can I get some questions for how to do capacity

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