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Finding maximum value

  1. Jan 28, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    For a,b belongs to R the maximum value of [itex](a-1)(b-1)+(1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/itex]


    2. Relevant equations

    3. The attempt at a solution
    I'm really clueless regarding how to start. So I started putting some arbitrary values but they were of no help. I can't even differentiate it as it is not a function in one variable. But it is clear that whatever value a and b attains, the following inequalities must be satisfied.

    1-a^2>0
    1-b^2>0

    This means that a and b must lie between -1 and 1. So far I could shorten the range of a and b. I also remember that I posted a very similar question earlier and someone suggested me to use geometric methods rather than solving it algebraically. OK, so I assume a function [itex]y=\sqrt{1-x^2}[/itex]. Now by inspection the original function transforms to [itex](x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)[/itex].

    But I really don't know how this helps me.
     
    Last edited: Jan 28, 2014
  2. jcsd
  3. Jan 28, 2014 #2
    Lagrangre? Your transformation is also not correct.
     
  4. Jan 28, 2014 #3
    You are proceeding in the right way! You have found out the range of 'a' and 'b'. Now this is just like finding the maximum of

    (a-1)(b-1) + (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) when a,bε[-1,1]

    For that maximize the first expression i.e, (a-1)(b-1) by substituting the appropriate value from the set[-1,1]; and maximize the second by i.e, (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) by first maximising (1-[itex]\sqrt{1-a^{2}}[/itex]) with the help of differentiation and doing the same for (1-[itex]\sqrt{1-b^{2}}[/itex]).

    Summarizing, all I want to say is, separately maximize each of the component by substituting numbers from [-1,1].

    Hope you get it. Best of luck!

    Regards.
    ADI
     
  5. Jan 28, 2014 #4
    Hint: do you think that for the maximum that a = b ? :D

    What does this mean for your equation?
     
  6. Jan 28, 2014 #5

    Ray Vickson

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    You should not claim 1-a^2 > 0 and 1-b^2 > 0; you just need 1 - a^2 >= 0 and 1-b^2 >= 0. That makes all the difference in the world: with strict inequalities you have an open set of feasible (a,b) values, and so there might not be a maximum at all. With non-strict inequalities you get a closed set, so are guaranteed the existence of a maximum (and minimum). Also, because you have inequalilty constraints on the variables you cannot be sure that differentiation will lead to a proper solution. (Hint: in this case it does not.)
     
  7. Jan 28, 2014 #6

    haruspex

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    No, Lagrange:wink:.
     
  8. Jan 28, 2014 #7

    haruspex

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    I fail to see how maximising the terms independently will work. Are you confusing [-1, 1] with {-1, 1}?
     
  9. Jan 28, 2014 #8

    haruspex

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    My immediate thought is to substitute trig functions. Should make the calculus a bit easier, but it doesn't help as much as I'd hoped.
     
  10. Jan 28, 2014 #9

    Ray Vickson

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    If you know about derivatives, the necessary conditions for a maximum of ##f(a,b)## = your function are:
    [tex] \frac{\partial f}{\partial a} = \begin{cases}
    0 &\text{ if } -1 < a < 1\\
    \leq 0 &\text{ if } a = -1 \\
    \geq 0 &\text{ if } a = +1
    \end{cases}[/tex]
    with the same type of conditions for ##b##. This is not very useful in itself, but it does act as a reminder to look not only at interior points, but also at the endpoints of the intervals for ##a## and ##b##.
     
  11. Jan 28, 2014 #10

    haruspex

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    The substitution of trig functions handles that automatically here, no?
     
  12. Jan 28, 2014 #11

    Ray Vickson

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    I don't see why it is easier to substitute ##\theta = \pi \text{ or } 0## instead of ##a = -1 \text{ or } 1##. Admittedly, writing ##a = \cos(\theta)## does restrict ##a## to the desired interval, but the endpoints still need checking manually.
     
  13. Jan 28, 2014 #12

    Ray Vickson

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    I don't see why it is easier to substitute ##\theta = 0 \text{ or } \pi## instead of ##a = -1 \text{ or } 1##.
     
  14. Jan 28, 2014 #13

    Ray Vickson

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    I don't see why it is easier to substitute ##\theta = 0 \text{ or } \pi## instead of ##a = -1 \text{ or } 1##. Admittedly, the substitution ##a =\cos(\theta)## does automatically restrict ##a## to the correct interval, but the endpoints still need to be checked manually.
     
  15. Jan 28, 2014 #14

    haruspex

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    Are you sure? That is not clear to me. If the max is at a = 1 then as theta passes through 0 the function will reach the max then fall off again, making it a local maximum.
     
  16. Jan 28, 2014 #15

    Ray Vickson

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    The function
    [tex] g(\theta,\phi) = (\cos(\theta)-1)(\cos(\phi)-1) + (1-\sin(\theta))(1-\sin(\phi))[/tex]
    respresents the original function
    [tex] f(a,b) = (a-1)(b-1) + (1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/tex]
    only for ##0 \leq \theta, \phi \leq \pi##. If ##\theta## or ##\phi## are a bit less than 0 or a bit more than ##\pi##, the "sin" parts go negative and so no longer represent the square roots in f.

    In this problem the partial derivatives of f are not both zero at the maximum, and the partial derivatives of g are also not both zero at the maximum. I don't want to say more now, because I don't want to give away the solution.
     
  17. Jan 29, 2014 #16

    utkarshakash

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    Your method is pretty much easier than others but I am not able to realise why the maximum should occur when a=b
     
  18. Jan 29, 2014 #17
    No, I am not. For finding the maximum value of two monotonous functions f(x) + g(x) we find the max value of each and then add them together, right? Since the values of a and b are confined to [-1,1], substituting the two extreme values i.e, -1 or 1 has a possibility of getting us the max value.
     
  19. Jan 29, 2014 #18

    Ray Vickson

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    The suggestion to take a = b is based on the symmetry of f(a,b). However, it is dangerous: just having symmetry does not imply that a = b at the solution of an optimization problem. Symmetry just implies that if (a,b) = (u,v) is a solution, then (a,b) = (v,u) is also a solution. This actually happens in this problem for the minimum of f! However, I won't say what happens for the maximum.
     
  20. Jan 29, 2014 #19

    haruspex

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    Wrong. If f and g reach their maxima for different values of x then max{f+g} < max{f}+max{g}.
     
  21. Jan 29, 2014 #20

    utkarshakash

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    So how should I start? You already said that trig substitution does not work.
     
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