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Maximizing Profits Problem

  1. Dec 31, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img707.imageshack.us/img707/8950/calculusisu.jpg [Broken]


    3. The attempt at a solution

    Looking for some guidance for this problem, I will add information as i go along.

    I think the steps are:
    1) Develop a formula that represents the cost
    2) Find critical points
    3) evauluate the critical points

    What i am unsure of is am i allowed to vary the length of the highway and gravel road? Im asked to maximize profit or i can minimize cost. Is this cost of operating or of construction as well? If its of operating only i can easily deduct the construction cost from the inital 7.8 million. Also if its of operating only, can i not say operate as fast as possible to reduce hours?

    Please advise
    thank you.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 31, 2009 #2

    LCKurtz

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    You might want to build the road at an angle and leave the highway sooner to take advantage of the shorter overall trip. Where to start cutting across through the forest is really the only variable you have to work with.

    Since profit = revenue - expenses and your revenue is fixed at the contract amount, minimizing cost and maximizing profit amount to the same thing.

    You have speed limits specified but, yes, given that restriction. And remember the cost of construction also depends on where you cut across.
     
    Last edited by a moderator: May 4, 2017
  4. Jan 2, 2010 #3
    I see,

    So in this case im allowed to alter the location of the gravel road, which can make it longer then 3Km?
     
  5. Jan 2, 2010 #4
    Im sorry, i asked because many people told me many different things, and its not anyones fault, many have blamed the question. Here are the 3 possible routes i have come up with..
    Im not sure which one the question wants me to use.

    http://img94.imageshack.us/img94/2023/lastscand.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Jan 2, 2010 #5

    Dick

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    I would go with option 3). Option 1) has a fixed cost, there is nothing to vary. Option 2) will always cost more than option 3) because the length of gravel is longer if you put the start of the gravel road in the same place.
     
  7. Jan 2, 2010 #6

    LCKurtz

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    Yes. The gravel road may be longer but the highway part is shorter and the total distance is less. But you must remember that it is not the total distance that matters. It is the total cost. For example, if the cost of building and traveling on the gravel was prohibitively expensive you might want the original route keeping the gravel part as short as possible even though the route is longer. And if, hypothetically and unrealistically, the gravel part was cheaper overall than the paved part, you would use a straight route avoiding the pavement entirely.

    That is why it is an optimization problem. Figure out where to cut across for maximum profit. Realize that, in principle, the answer could come out at either end or in the middle.
     
  8. Jan 2, 2010 #7
    Thank you very much guys,

    I did as you have instructed and came up with this:

    http://img51.imageshack.us/img51/4687/lastscancw.jpg [Broken]


    After minimizing the cost i found (a) to be 1.3865 Km.

    If someone would confirm, id be indebted. :) If not, i udnerstand since its a lengthy process.

    Ill keep my foruma for cost private for now unless someone wants to see it...

    Cost = construstion Cost + Operation cost

    ^ this is what i minimized in terms of (a) for both operation and construction.


    In addition, If you notice This method above, takes care of the possibilites 1 and 3 in post # 4.
     
    Last edited by a moderator: May 4, 2017
  9. Jan 2, 2010 #8

    LCKurtz

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    I got 1.334 for a.
     
  10. Jan 3, 2010 #9
    I shall retry and have a friend look over my work, then will post back. It may be that our cost formulas are different.
    Thank you for taking the time.
     
  11. Jan 3, 2010 #10
    Just gave it another try and i still get a = 1.3865

    Here is my formula for cost:

    Cost Total = (2 000 000 – 200 000a + 500 000(√(9+a^2 ))) + 300 000 [((50-5a)/7 + (13/8*(√(9+a^2 ))]


    My derivative simplifies to:

    C' = -2 900 000/7 + (987 500a)/√(9+a^2 )

    Also, im in need of some good free graphing calculator website i can use to plot that finction if anyone knows of one.
     
    Last edited: Jan 3, 2010
  12. Jan 3, 2010 #11
    I just compared my value of 1.3865 with 1.334 in terms of cost and i get a lower cost with my value. But this could be due to some difference in the formulas we developed. Please advise.

    Thank you sir!
     
  13. Jan 3, 2010 #12

    LCKurtz

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    You have one too many zeroes in your 300000 term. 100*300 = 30000 trips. Then you need to remember they are round trips so it should actually be 60000.
     
  14. Jan 3, 2010 #13
    I agree they are round trips, 100 of them.

    Round trip:
    A trip from one place to another and back.


    100 * 2 = 200 one way trips in 1 day

    200 trips/day * 300 days / year *5 years

    I get 300 000 one way trips in 5 years. That is, 150 000 times going there and 150 000 going back.

    I do get 60 000 per year then * 5 years. (since operation is for 5 years)
    Im not sure if im missing something about the meaning of round trips. The deninition above is the one i know of.

    Once again I appreciate your time and help. Thank you.
     
    Last edited: Jan 3, 2010
  15. Jan 3, 2010 #14

    LCKurtz

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    Yes, you are exactly correct. Neither of us were confused about the round trips but I missed the part about it being 5 years.

    Just type online grapher into google for some places to get a graph.
     
  16. Jan 3, 2010 #15
    Ok worked out well. Thank you very much for your help. I appreciate it.
     
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