Math

posted by VIctoria

P=20x+30y

subject to:
2x+3y>30
2x+y<26
-6x+5y<50
x,y >0

  1. bobpursley

    Two ways:
    Graphical...
    plot the contraint areas 2x+3y=30
    2x+y=26
    -6x+5y=50
    x,y >0
    Now we have a nice theorem that states the max, and min, will occur on the boundry where constraint lines meet. So test the profit function 20x+30y at each corner, and you will find the max.

    Second method: Analytical
    http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue

  2. Reiny

    looks like linear programming.

    let's look at each region.
    2x + 3y > 30
    the boundary is 2x + 3y = 30
    when x = 0, y = 10
    when y = 0 , x = 15 , so we have (0,10) and (15,0)
    plot these points on the axes and draw a dotted line,
    shade in the region above this line

    2x+y < 26
    repeat my method, shade in the region below the line 2x+y = 26

    careful with the last one, I will do that one as well.
    -6x+y < 50 -----> y < 6x+50
    draw the boundary line y = 6x+50
    when x=0, y = 50 , when y = 0 x = -25/3
    so your two points are (0,50) , (-25/3,0)
    shade in the region below that line

    Your boundary lines should look like this
    http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y+%3D+30,+2x%2By+%3D+26,+y+%3D+6x%2B50

    and here the actual intersection of the shaded regions.
    http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y%3E+30,+2x%2By+%3C+26,+y+%3C+6x%2B50

    notice that P = 20x + 30y has the same slope as 2x + 3y = 30
    so the farthest point to the right of your region will be the solution.
    Solve the corresponding equations to find that point, plug into
    P = 20x + 30y

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