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Optimization and Related rates

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined deposits on the line joining them, at a distance x from one stack, is given by the following equation, where k is a positive constants that depends on the quantity of smoke each stack is emitting.

    [itex]S=((64k)/(x^2)) + (k/((20-x)^2))[/itex]


    2. Relevant equations



    3. The attempt at a solution
    1st I take the derivative

    [itex]((-128k(20-x)^3))+2kx^3/(x^3)(20-x)^3))[/itex]

    Then i look for critical points

    x^3=0
    x=0

    20-x=0
    x=20

    The third one is where i get stuck

    (-128k(20-x)^3))+2kx^3=0
    k(-128(-x^3 + 60x^2 -1200x +8000) + 2x^3)=0
    (-128(-x^3 + 60x^2 -1200x +8000) + 2x^3)=0
    128x^3-7680x^2 +153600x - 1024000 + 2x^3=0
    130x^3-7680x^2 +153600x - 1024000=0
    130x^3-7680x^2 +153600x = 1024000

    how would I solve for this? It's a bit embarrassing on my part, but I never ran into a situation before where I had to find the root of a cubic function.
     
  2. jcsd
  3. Oct 17, 2012 #2

    Mark44

    Staff: Mentor

    We usually need to tell people that they need more parentheses, but here you can get by with fewer of them.

    S = 64k/x2 + k/(20 - x)2

    Since you're going to differentiate this, it's better to write it using negative exponents.

    S = 64kx-2 + k(20 - x)-2
    Even though you have lots of parentheses, you are missing some where they are needed. The numerator is -128k(20 - x)3 + 2kx3. The numerator should have a pair of parentheses around it.
    You can divide both sides by 2k to simplify this a little bit.

    -64(20 - x)3 + x3 = 0

    If you expand this, and then attempt to factor it, you might get lucky and find a rational root, or you might not. Although cubics can be factored, it's not a trivial matter to do so. If the exact root is not required, you can estimate it by looking at a graph or by using approximation techniques, such as Newton's method.
     
  4. Oct 17, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    First: divide through by 10, to make it very slightly easier:
    p(x) = 13 x^3 - 768 x^2 + 15360 x - 102400 = 0.

    As others have pointed out, there are exact formulas for solving such problems, but they are messy. Often, it is better to just ignore those formulas and proceed more directly. You can draw a rough graph of p(x) to get a feeling for where a root might be located, or you can use a "bisecting" method: p(0) = -102400, p(100) = 6753600, so there is a root between 0 and 100 (because p changes sign between 0 and 100). Look at x = 50: p(50) = 370600. There is a root between 0 and 50. So, try x = 25: p(25) = 4725. There is a root between 0 and 25. Look at x = 12 (roughly half of the current interval): p(12) = -6207. So, there is a root between 12 and 25, etc. A couple of additional steps of that type will get you very near the root, perhaps even to its exact value.

    Note: you could try to speed up the process, by using something called the "Secant Method". Basically, this approximates p(x) by a straight line joining two points on its graph, and looks for a root of the straight line. For example, the line through (12,p(12)) and (25,p(25)) has equation y = p(12) + [(p(25)-p(12))/(25-12)]*(x-12) = 841*x - 16300, which gives zero at x = 16300/841 ≈ 19.4 . In this case, though, bisection is easier and just as fast, or faster.

    Of course, the classical way is Newton's method, but this often needs a reasonably close approximation to the root to get going properly. In this case, a few bisection steps should lead to a reasonable starting point.

    Finally, there is the modern method, which is to use a good scientific hand-held calculator, or a computer package such as the EXCEL Solver tool or Matlab, or to submit the problem to the free, on-line package Wolfram Alpha.

    RGV
     
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