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Optimizing Values Question

  1. Dec 8, 2005 #1
    A constructiuon company has been offered to build-operate contract for 7.8million to construct and operate a trucking route for five years to transport ore from a mine site to a smelter. the smelteris located down a major highway and the mine is 3km into the bush off the highway. the smelter is 10 km down the highway from the mine side road.

    to upgrade highway(repave)will be $200000/km
    new gravel from mine to highway is $500000/km

    operating costs
    100 return trips each day for 300 days for the five years
    to drive on the gravel road it will be $65/h and average speed of 40km/h
    to drive on the highway it will be $50/h and average speed of 70km/h

    use calculus to determine if the company will accept the contract and the distances of paved and gravel road producing optimumconditions(max profit). what is the maximum profit? do not consider the time value of money in your claculations.

    ^^^^that was the hole question^^^^^^

    ok so all i have so far is the start of an equation to figure out the total cost . and once i have that i would differentiate that and find at what value x=0 to find best conditions.right......?

    i will enter all the work i have later when i get home cause im still workin on it and my school comps are not that fast so i will wait if there is any helpful instie that anybody can offer it would be greatlt appretiated.
  2. jcsd
  3. Dec 8, 2005 #2
    my equations so far is
    C(x)= (1500000)+(200000x)+[((13-x)19500000)/40]+(15000000x/70)
    and if i find the derivative of that it doesnt make any sense so if i have any more revalations i will share but i dont really know
  4. Dec 8, 2005 #3


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    That doesn't tell me anything. What does x represent? Where does "13-x" come from? Are you assuming that the new "mine road" runs perpendicularly to the paved road?
  5. Dec 8, 2005 #4
    x is the distance of road that is paved and 13-x is the total distance minus x is what is left over and is not paved(gravel) and yes the road runs perpendicular to the highway
  6. Dec 8, 2005 #5


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    This ugly nightmare reminds me of my youth. I started college pursuing a degree in mining engineering back in 1976.

  7. Dec 9, 2005 #6
    well can ya help any?
  8. Dec 9, 2005 #7


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    If someone put a gun to my head and made me solve this problem, the first thing I'd do would be to write a computer program to model the situation. Then, from the result of the computer program, I'd write up the "calculus" proof. The purpose of the computer program is to provide a peek at the answer in the back of the book, more or less. Furthermore, this is how an engineer in the field would approach the problem, not from a point of view of a pure application of calculus.

    There are just so many opportunities for making an error in setting up the problem in calculus. With a computer program, you can have it spit out the costs for the various parts of the program and you can verify that these are correct and add up by eye.

  9. Dec 10, 2005 #8
    well thanks CralB but unfortunatly i dont have any computer prgrams that will help me.and also the question has changed, there is no perpendicular road from the mine to the highway i now have to find out at what angle does a road from the mine intersect the highway that makes the shortest distance.and also figure out how much i must pave to get highest profit.
    Last edited: Dec 10, 2005
  10. Dec 10, 2005 #9


    Staff: Mentor

    You can approach it one of two ways that I can think of immediately. Either you can write the total cost as a function of the angle your dirt road makes with the main road, or you can write the total cost as a function of the length of the dirt road. I would probably write it as a function of the angle.

    Once you have that you just optimize like normal (differentiate, find zeros, and make sure it is a max).

  11. Dec 10, 2005 #10


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    That's good. It's a much more tractable problem for calculus now. I wouldn't bother with the computer program with this one.

  12. Dec 11, 2005 #11
    ok ive made some real prgress so far but i need help finding x:

  13. Dec 11, 2005 #12


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    WHY are you assuming that the mine road runs perpendicular to the paved road? If that is the case then you have no problem: the distances and therefore the cost are fixed.

    Instead, assume that the mine road meets the paved road at distance "x" from where the perpendicular would meet it. Since the mine is 3 km from the paved road (perpendicularly) the mine road is the hypotenuse of a right triangle with legs of length 3 and x: the length of the mine road is [itex]\sqrt{9+ x^2}= (9+ x^2}^{\frac{1}{2}}[/itex]. The length of the paved road would be 10- x and the total cost to construct and repave the roads is
    [tex]500000(9+ x^2)^{\frac{1}{2}}+ 200000(10-x)[/itex]

    I assume that "100 return trips each day for 300 days for the five years" means 300 days each year for 5 years: 100(300)(5)= 150000 trips. The cost of a trip over the mine road would be [itex]2600(9+x^2)^{\frac{1}{2}}[/itex] and the cost of a trip over the paved road [itex]3500(10-x)[/itex] Since each trip is a round trip that will be 300000 trips in 5 years.
    The total cost in 5 years is
    [tex]500000(9+ x^2)^{\frac{1}{2}}+ 200000(10-x)+ 300000(2600(9+x^2)^{\frac{1}{2}}+ 3500(10-x))[/itex]
    Find the value of x that minimizes that, find that minimum cost and compare it to the amount the company is being paid.
  14. Dec 11, 2005 #13
    ok i understand all that but where do u get 2600 and 3500 from?

    and it also doesnot take into concideration that the road from the mine to the high way could be paved

    and my teacher had originally said that the road was perpendicular to the highway anf ur right that caused no problems but when i told him that that made it too easy he changed it.
    Last edited: Dec 11, 2005
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