1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic math Olympiad problem

  1. May 6, 2017 #1
    <Moderator's note: moved from a technical forum, so homework template missing.>
    upload_2017-5-6_22-11-44.png
    so this is the question.
    i want to know if there is a solution without using calculus maybe trig substitution maybe other methods?
    i tried trig substitition
    i let u = √2 cosx
    and v be sinx
    am i on the right track
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. May 6, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You need two variables - with only x you cannot cover the whole plane you have to search.
     
  4. May 8, 2017 #3
    ok fine let
    v be siny
    but this only makes the problem complicated how am i supposed to minimum value of function with different variables
     
  5. May 8, 2017 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It has to be a minimum with respect to both variables. While that is technically not sufficient to have a global minimum, it will do the job here.
     
  6. May 8, 2017 #5
    so terms containing x should be minimum themselves and terms containing y must be local minimum too?
     
  7. May 8, 2017 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Yes, or the same with u and v without substitution. If a point is not a minimum with respect to these variables, there is a point nearby which has a lower value.
     
  8. May 8, 2017 #7
    Are you allowed to use a calculator??
     
  9. May 9, 2017 #8
  10. May 9, 2017 #9
    @mfb
    i am just realizing i got back to using calculus and partial derivative i know that method already is there any other
     
  11. May 9, 2017 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Sometimes there are other options. You have the sum of two squares. Both squares are not negative, so if there is a point where both squares are zero, or at least at a global minimum within the allowed parameter range, you found a global minimum. Then you just have to check if it is within the given range of the coordinates.
    There is no such point in this problem (unless you allow imaginary arguments).
     
  12. May 10, 2017 #11
    so is the calculus the only method? without imaginary numbers only real because i have it done the calculus way already
     
  13. May 10, 2017 #12
    Did you try AM-GM ?
     
  14. May 12, 2017 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    One thing that makes it hard is having both variables in both terms. If you were to make a substitution to two new variables, x and y, so that one term only contains x then you can concentrate first on minimising the other term wrt y.
    The downside is that checking the given bounds gets awkward.

    Another possible start is to consider what happens on the boundaries.
     
    Last edited: May 12, 2017
  15. May 13, 2017 #14
    sorry for the late reply

    @Buffu
    how to apply am-gm here

    @haruspex
    yup i am finally seeing that calculus is the most straight forward approach here
     
  16. May 13, 2017 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Standard differential calculus approach will lead to quartics to solve.
    I think I have the answer, using my first suggestion. Write x+u for v and find the min wrt u. That should give you an expression for v in terms of u. It is rather nasty, but see if you can spot a fairly simple pair of values satisfying it.
    Showing that produces the minimum is easier than finding it from calculus.
     
  17. May 13, 2017 #16
  18. May 13, 2017 #17
    ##\displaystyle (u - v)^2 + \left(\sqrt{2- u^2} - {9\over v}\right)^2 \ge 2(u-v)\left(\sqrt{2- u^2} - {9\over v}\right)##

    Maybe do ##v = ku## now.

    I don't think it simplifies the equation much but better than nothing, right ?
     
  19. May 13, 2017 #18
    okay i will give that a try too give me some time i am now doing another problem
     
  20. May 13, 2017 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Very neat, and yes, that is the answer I got.
    The method at that link does produce quartics - see the expression with both y and 1/y3 - but it turns out to factorise nicely.
     
  21. May 13, 2017 #20
    oh if thats what you meant by quartics then yes but anyways i will give your method a try too
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Basic math Olympiad problem
  1. Olympiad problem (Replies: 7)

  2. Basic Trig Problem (Replies: 4)

Loading...