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Reply to an old thread on the ratio of parabolas

  1. Jul 30, 2013 #1
    I was searching through old threads that had not been replied to and I found one by 'japam' from Feb 6, 2005. After solving it, I realized that the thread was closed so I am reposting it now in case 'japam' is still curious :)

    --------------------------
    His question was:
    Suppose you have two quadratic functions: y1 = ax^2+bx+c and y2 = dx^2+ex+f. Suppose that the first one is concave up with its minimum above the x-axis and the second one is concave down with its maximum also above the x-axis. Furthermore, suppose that the two functions intersect at two points that pass through the straight line y3 = gx+h.

    My question is
    Could the maximum of (dx^2+ex+f)/(ax^2+bx+c), be splitted as the
    max( (dx^2+ex+f)/(gx+h))*max((gx+h)/(ax^2+bx+c)) ?

    I don't know how to put drawings here, but i hope the argument has been clear to understand.
    I was trying some numerical examples on my PC and the results were positive, but i don't know the general proof. Thanks for your comments.
    --------------------------

    I tried out a handful of examples on my trusty TI-86. I believe that this is indeed true, that the maximum of the rational expression can be divided into the product of the individual maxima as written above. This follows from the following statement which I was able to prove just now but I don't know if it's a formal theorem:

    Let f(x) and g(x) be two parabolas that pass through the points (a,b) and (c,d), where a≠c and bd > 0. Let h(x) be the line that passes through the points (a,b) and (c,d). Then the unique relative extremum of f(x)/h(x) and g(x)/h(x) on the interval a < x < c both occur at the same point x = f, where a < f < c. In other words, all quadratics which pass through two given points in the plane have the property that the maximum value of the function given by the ratio of the quadratic to the line between the two points occurs at the same x-coordinate regardless of the quadratic function chosen, as long as it passes through the two given points.

    Applied to the question posted by 'japam' about 17/2 years ago, we see that this means that the x-coordinate which maximizes the ratio of the quadratics is the same x-coordinate that maximizes each individual ratio in the product that was mentioned.

    For those who are interested, here is my proof:
    Let f(x) be a parabola that passes through the points (a,b) and (c,d). Let h(x) = mx + e be the equation of the line through the two points (a,b) and (c,d). Then the family of parabolas f(x) has only one degree of freedom as three points determine a parabola in the plane. I will call this single degree of freedom, r.
    Then it follows that f(x) = r(x-a)(c-x) + mx + e is a parabola which satisfies the conditions mentioned above. Then f(x)/h(x) = (r(x-a)(c-x))/(mx+e) + 1. The derivative of this function is:
    (f(x)/h(x))' = ((mx+e)(r)(-2x+a+c)-r(x-a)(c-x)(m))/(mx+e)^2
    Since f(a)/h(a) = f(c)/h(c) = 1 and f(x)/h(x) is continuous on [a,c] and differentiable on (a,c) it follows by Rolle's Theorem that there exists at least one point in (a,c) such that the derivative equals zero. Since the parabola f(x) is either always above the line or below the line h(x) and does not change concavity, it follows that there is a unique relative extremum on the interval.
    Setting the numerator of the above derivative equal to zero and dividing both sides by r we get:
    ((mx+e)(-2x+a+c)-(x-a)(c-x)(m))/(mx+e)^2 = 0.
    However, this means that the solution for x does not depend on the degree of freedom, r.
    Thus, all parabolas through (a,b) and (c,d) will have a maximum ratio with the line h(x) at the same x-coordinate.

    Junaid Mansuri
     
    Last edited: Jul 30, 2013
  2. jcsd
  3. Jul 30, 2013 #2

    mfb

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    Staff: Mentor

    I guess japam won't read this thread, but it is an interesting theorem.

    You can simplify it a bit:
    A maximum/minimum of f(x)/h(x) is also a maximum/minimum of (f(x)/h(x)-1)/r (as r is does not depend on x), and this is independent of r. Therefore, the x-value cannot depend on r.

    The minimum has an analytic solution (I replaced e by f, otherwise WA interprets it as Euler's number).
     
  4. Aug 2, 2013 #3
    Thanks for the observation. That's a nicer way of looking at it.
     
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