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Thread: Minimizing Costs Problem

  1. #1
    Jun 2010

    Exclamation Minimizing Costs Problem

    Hi, I'm currently in Pre-Calculus in High school, a course prepping for university; I'm hoping this thread is in the appropriate section! I apologize for the long question, but I don't want to confuse anyone who is willing to help me.

    We are currently doing a project on minimizing the cost for a company that has been contracted to lay pipeline through a forested area between a natural gas well and a storage facility.

    Picture a rectangle where the horizontal is the roadway, the vertical is always 3000 m long, and in the middle is where a forest is. There's a diagonal pipeline starting from the lower left corner of the rectangle (the well site) to the upper right corner of the rectangle (the storage facility). It wil cost $230/m to lay pipeline parallel to the roadway (horizontal line) and $260/m to lay pipeline through the forested area (diagonal line).

    However the diagonal line isn't directly connected from the well site to the storage facility--the company would lay 500 m of pipeline parallel to the roadway while the rest of the pipeline through the forested area (diagonal line). The horizontal line begins at 10 000 m, but decreases every 500 m as the pipeline parallel to the horizontal increases every 500 m.

    First I designed a spreadsheet with values. The first column (a) is 'trial number' [let's call this y], the second (b) is 'distance of pipeline parallel to roadway in meters' [let's call this x], third is 'cost of pipeline through forested area in $' (c), fourth (d) is 'cost of pipeline through forested area in $' fifth (e) is 'cost of pipeline through forested area in $' and last (f) is 'total cost of pipeline in $'. For example one of the rows would be:
    1 // 500y // √[((10 000 - x)) + (3000)] // {√[((10 000 - x)) + (3000)]} 260 // x 230 // [{√[((10 000 - x)) + (3000)]} 260] + [x 230]

    Simply put it would look like:
    1 // 500a // √[((10 000 - x)) + (3000)] // 260 c // 230 x // d + e

    I have to do this for about 25 trials until the distance of pipeline parallel to roadway is 10 000 m. I found that the least expensive path was when the pipeline parallel to the roadway is 4 500 m.

    I have to find the formula first for the equation using technology (I haven't gotten that far yet but I know what to do), but once I find the formula my next task is to find the least expensive path using calculus and algebra.

    My question is: I'm not exactly sure how to set this question up. Basically I have a rectangle where the vertical is always 3000 m. I have a diagonal line attached to another horizontal line; the horizontal line increasing 500 m while the horizontal line of the rectangle decreases 500 m. I know my solution would probably have to end up being 4 500.

    I'm not asking for a full solution (if you're not comfortable with that), but I would be very grateful for any help or tips on solving this.

    Thank you very much and again, I'm sorry that this question is long!
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  2. #2
    Member mfetch22's Avatar
    Feb 2010
    Columbus, Ohio, USA
    I'm kinda confused by your problem, if you have a picture to go along with it that would be great. I dont know the exacts of your problem but I know what type of problem this is: Optimization/Minimalization. Heres what you need to do:

    1) Set up two equations with the things you know in it, and the exact amount of piping of a certain price as x and y variables. Make sure one of the equations is for how much the total cost will be, the other will probably be for something like how much piping they have for something of a more geometric meaning. Make sure both equations are in terms of x and y, whatever x and y may be (probably the amount of piping for 230 and for 260)

    2) In the equation that isnt giving the total cost, move the variables around so one variable is on one side all by itself. This is equating one variable in terms of another, so you might have $\displaystyle x = 2y + 6$ or something like that, thats just an example.

    3) Take this expression for your one variable and plug it into the cost equation, you should realize that now your cost equation is all in terms of one variable.

    4) Graph the cost equation now in one variable on your calulator, and than just look for the point where the curve on the screen is the lowest, this represents the x value at which cost is the least. This x value is your x answer, just plug it into the other equation to find your y and now you have your answers of how much pipe of each type is needed.

    Hope this helps
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