SOLUTION: A barge owned by " Ernest Transport Co" travels between England & France, and can carry up to 24 cars or up to 8 buses. Each bus takes up the space of 3 cars. The owner matthew, n

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A barge owned by " Ernest Transport Co" travels between England & France, and can carry up to 24 cars or up to 8 buses. Each bus takes up the space of 3 cars. The owner matthew, n      Log On


   

Question 12593: A barge owned by " Ernest Transport Co" travels between England & France, and can carry up to 24 cars or up to 8 buses. Each bus takes up the space of 3 cars.
The owner matthew, needs you to determine the most profitable combination of cars & buses which can be carried on a particular trip. The following conditions apply.
a) On any trip at least 8 cars are to be carried; and
b) More than 3 buses are to be carried
c) Matthew requires a maximum income. He is being paid $300 per bus & $50 per car.

1) Translate this situation into a set of mathemathical relationships.
2) Draw graphs of the relationships.
3) Find region of overlap.
Could you please help me solve this problem.
Thanking you in advance.
Julie
Answer by AdolphousC(70) About Me  (Show Source):
You can put this solution on YOUR website!
Julie, First you need to consider your constraints, and these are easy, as they are just 1 variable contraints,
  • On any given trip the barge can carry anywhere from 8 to 24 cars
  • On any given trip the barge can carry anywhere from 3 to 8 buses
    Now, lets make the number of cars your x-axis, and the number of buses your y-axis.


  • Graph the simple car constraints first, if the barge carries 8 cars, you will have a vertical line at X = 8, since this in the minimum number of cars the barge can carry. ( Shade right )
  • Now graph the maximum, X = 24 ( Shade Left )


    I can't graph vertical lines on here, but you should now have two vertical lines, one at x = 8 and one at x = 24.


  • No graph your simple bus contraints, if the barge carries the minimum buses you will have a horizontal line at y = 3 ( Shade Up )
  • If the barge carries the maximum busses you will have a horizontal line at y = 8 ( Shade Down )


    Wait, we're not finished with the constraints yet, you should have a box in the middle of your graph that is shaded with the corner points being (8,3),(24,3),(8,8) and (24,8)


    Now the last constraint, here you will need to find and plot two points, and use your slope formula m=+%28Y2-Y1%29%2F%28X2-X1%29 to find your equation. if the barge carries no cars, then it can carry 8 buses, so your first point on your line is going to be (0,8)


    Now this part is tricky, and we're going to go outside of the contraints to find out line. If the barge were to carry only 1 bus, we know the barge would fit 21 cars. So the next point of your line is ( 21, 1 )


    Use these two points (0,8) & (21,1) to find the equation of your last constraint which will be y=%28-1%2F3%29x and graph this contraint on your paper, shading down.


    Now here is the easy part, move around the outside edge of your figure that is shaded, and get all the "Corner Points" these are the only options you have for a maximum profit.


    Using these corner points, plug them into your profit equation which is...
    P+=+300B+%2B+50C+ which ever set of points gives you the largest P value will be your answer for the problem, and tell you how many cars and how many buses should be transported in order to get the most money.