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Homework Help: Complex Optimization Problem regarding Areas

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Part 1:
    A forest in the shape of a 50km x 50 km square has firebreaks in rectangular strips 50km by 0.01 km. The trees between two fire breaks are called a stand of trees. All firebreaks in this forest are parallel to each other and to one edge of the forest, with the first firebreak at the edge of the forest. The firebreaks are evenly spaced throughout the forest. The total area lost in the case of a fire is the area of the stand of trees where the fire started and the combined area of the firebreaks.

    Part 2:
    Now suppose that the firebreaks are arranged in two equally spaced sets of parallel lines, as shown in Fig 2. The forest is again 50km x 50km and each break is 50km x 0.01km. Find the optimal number of firebreaks in each direction to minimize the area lost due to the fire. Assume that there must be the same number of firebreaks in each direction. (eg. 3 North to South and 3 East to West). Make sure you do not count the area of overlap in the firebreaks more than once. Assume that you know the critical number is 5000 for part 2 in order to find the value of n we are interested in.

    The Figures: http://i42.tinypic.com/3324bk3.jpg
    2. Relevant equations

    Here are the equations I have came up with.
    Domain for n is [1,5000]

    Area of Burned Forest = Af(n) = ((50/n)-0.01)^2
    Area of Firebreaks = Ab(n) = 2n^2((1/2n)-0.0001)-0.0001n^2
    Area saved = As(n) = 2500 - [Af(n) + Ab(n)]
    d/dx As = 5000-n-(n^4)/5000+2n^3

    3. The attempt at a solution
    What I've been trying to do is find the critical numbers of the equation from the derivative.
    Our professor gave the the critical number as 5000, but 5000 is the max amount of fire breaks you can have, as 5000x0.01km = 50km. I know the solution is 17 firebreaks from a maximum of 17.09 I obtained from a graph. I just cannot seem to get there with the math.
    When I emailed my professor he told me:
    I simplified the equations but the values I'm getting are still outside of the domain except for 5000. I can prove that 5000 is a minimum but I don't see how that leads to a solution.

    I have solved part 1, I'm just stuck at part 2.

    Links to my work if the text is hard to read.
    http://i39.tinypic.com/qxq4ww.jpg - Part 2a
    http://i41.tinypic.com/16le9fq.jpg - Part2b
    Last edited: May 15, 2010
  2. jcsd
  3. Nov 28, 2010 #2
    I know you can't use this now as you would be way past due :redface:, but here's to help anyone else who has seen the same problem.
    I came up with a different equation:
    A1 burned area(n)=((50/.5n)-.01)2=((100/n)-.01)2=10000n-2-2n-1+.0001 Because half of the firebreaks are n/s and the other half are e/w, you account for .5 of the total firebreaks for each side of burnout areas.
    Atotal firebreaks(n)=0.5n-.0001(0.5n)2=.5n-.000025x2 Take the area of 1 firebreak (.5km2) minus the overlap areas (.0001km2). The overlap areas are only subtracted from the verticle OR horizontal firebreaks, not both.
    Asaved=2500-(A1 burned area+Atotal firebreaks) This function gives you total firebreaks (x axis) vs total area saved (y axis). Either graph it out and find the max button on your calculator, or find the derivative and set to 0. I will break down the derivative.
    d/dn (2500-10000n-2+2n-1-.0001.-.5n+.000025n2)=0
    20000n-3-2n-2-.5+.00005n=0, x=34.1995
    Therefore there are 17 north/south firebreaks, and 17 east/west firebreaks.

    I hope this helped anyone looking for an answer to this problem, and any feedback would be welcomed.
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