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Homework Help: A variational problem with the constraint that the function be decreasing

  1. May 15, 2012 #1
    The problem is in the attached document. Thanks for any suggestions!
    Lennart
     
  2. jcsd
  3. May 16, 2012 #2
    The mathematical problem:

    $\theta$ is a constant that equals 0.8.

    We consider the set

    '$ S=\left\{(F,h):F is a decreasing function from R^{+} to R^{+}, h\in R, 0=1- \frac{\theta+1)}{\theta} \frac {(\int^{h}_{y=0} F(y) dy)}{/F(0)} \frac{F(0)-\frac{1}{2}}{F(0)-F(h)} \right\}$'
    The function L is defined on S by
    $ L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h$
    We want to find the maximal value of L.

    A further problem:
    Denote by $L(\theta)$ the maximal value of L. What is $max_{\theta \in (0,1)} \frac {L(\theta)}{\theta}$ ?
     
    Last edited: May 16, 2012
  4. May 16, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    If you want LaTeX to be readable in this forum, you need to remove the $ delimiters and enclose your commands between "[tex ]" and "[/tex ]" delimiters (no space after the word tex--I used a space so as to not confuse the system). For example, here is one of your formulas (justa copied and pasted between the delimiters I:
    [tex] L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h[/tex]

    RGV
     
  5. May 16, 2012 #4
    [tex]\theta[/tex] is a positive constant.

    We consider the set

    [tex] S=\left\{(F,h):F is a decreasing function from R^{+} to R^{+}, h\in R, 0=1- \frac{\theta+1}{\theta} \frac {\int^{h}_{y=0} F(y) dy}{F(0)} \frac{F(0)-\frac{1}{2}F(h)}{F(0)-F(h)} \right\}[/tex]
    The function L is defined on S by
    [tex] L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h[/tex]
    We want to find the maximal value of L, denoted by [tex]L(\theta)[/tex].

    A special question:
    What is L(0.8) ?

    A general question:
    What is [tex]max_{\theta \in (0,1)} \frac {L(\theta)}{\theta}[/tex] ?
     
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