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Maximum of a function

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data
    I have an equation q*(p-q). I need a function with which I could specify a p, and which would ouput a q at which q*(p-q) is maximized.

    For example:

    f(p) -> analyzes y = q*(p-q) -> outputs the q at which y reaches it's maximum (in other words, f(p) = maximum of y at a given p).

    2. Relevant equations
    q*(p-q)


    3. The attempt at a solution

    Tried to apply the second partial derivative test to this, but it came out inconclusive with a D = 0. It seems like I applied it where it isn't needed at all.
     
  2. jcsd
  3. Feb 12, 2012 #2

    tiny-tim

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    Hi Cinitiator! :smile:
    shouldn't do :confused:

    (did you d/dp instead of d/dq ?)

    show us what you did :smile:
     
  4. Feb 12, 2012 #3

    SammyS

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    First of all, (and I expect that you will see this mentioned several times in this thread)
    "I have an equation q*(p-q)."​
    " q*(p-q) " is a mathematical expression. It's not an equation. An equation has two mathematical expressions separated by an " = " sign.

    You say that for any particular value of p, you want to find a q which maximizes q*(p-q) .
    Let g(q) = q*(p-q). That is equivalent to: g(q) = -q - pq .

    That's a quadratic in q, with q-intercepts of q=0, and q=p. The vertex occurs at q = p/2. q2 has a negative coefficient, so that should tell you whether the vertex corresponds to a maximum or a minimum.
     
  5. Feb 12, 2012 #4
    Why not to use partial derivatives instead?

    One could do this:

    1) Find the partial derivative with the respect to q
    d/dq [q(p-q)] = d/dq [q*p-q^2] = d/dq [q*p ] - d/dq [q^2] = p - 2q

    2) Then find critical points
    d/dq = 0
    p - 2q = 0
    (-2)q = -p
    q = (-p)/(-2)

    q = p/2


    Then f(p) = p/2

    I tested it graphically and it seems to work.


    I looked through my notes, and it seems that this was the case, lol. I inverted p and q multiple times. Wish I had a better working memory.
     
  6. Feb 12, 2012 #5

    tiny-tim

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    Hi Cinitiator! :smile:
    yup! :smile:

    (btw, with quadratics, you can find the turning-points without calculus, if you complete the square :wink:)
    you'll have to …

    mind your p's and q's! :smile:

    :biggrin: Woohoo! :biggrin:
     
  7. Feb 12, 2012 #6

    SammyS

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    [STRIKE]Nice[/STRIKE] Great reply tim !

    Too bad this Forum doesn't let us rate replies !
     
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