Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hard logarithmic/implicit differentiation question.

  1. Oct 26, 2010 #1
    QUESTION HAS BEEN SOLVED ! :) THANKS MARK!

    1. The problem statement, all variables and given/known data
    Shark Inc. has determined that demand for its newest netbook model is given by [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/80/fa9842dd3479ee1baaab655e02a1c41.png, [Broken] where q is the number of netbooks Shark can sell at a price of p dollars per unit. Shark has determined that this model is valid for prices [tex]p \geq 100[/tex]. You may find it useful in this problem to know that elasticity of demand is defined to be [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/c9/9d605854cd629ac16035b6765c1d051.png [Broken]

    1) Find E(p), your answer should be in terms of p only.
    2) What price will maximize revenue, if the price is less than 100, write NA

    2. The attempt at a solution

    My answer for #1 (which was wrong) using the formula was:

    [tex]-[(e^(3*ln(p)-0.002p+7)*(3/p-0.002))]*(\frac{p}{(e^(3*ln(p)-0.002p+7)})[/tex]

    First i solved for q. As all answers must be in terms of p.

    Isolating i got:

    [tex] ln(q) = 3ln(p) - 0.002p +7 [/tex]

    [tex] q = e^\left(3ln(p) - 0.002p +7\right) [/tex]

    Then I differentiated using the chain and logarithm rules

    [tex] q' = \left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right) [/tex]

    So when I put these values into: [tex] E\left(p\right) = -q' * \frac{p}{q} [/tex]

    I got

    [tex] E\left(p\right) = -1 * \left[\left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right)\right] * \frac{p}{e^\left(3ln(p) - 0.002p +7\right)} [/tex]

    But this is wrong, why?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 26, 2010 #2

    Mark44

    Staff: Mentor

    This looks fine. You might want to simplify it, though, to e7*p3*e-0.002p.
    Instead of doing it this way, why don't you differentiate the original equation implicitly. I think that would be simpler. When you're ready to put your answer for E(p), you can replace q using the formula you got above, so that E(p) is written in terms of p alone.
     
    Last edited by a moderator: May 5, 2017
  4. Oct 26, 2010 #3
    is q' wrong?

    I did it implicitly using the original equation and got the same answer.

    Implicit

    [tex](ln(q)-3ln(p)-0.002p)' = (7)' [/tex]

    =

    [tex]\frac{q'}{q} - \frac{3}{p} - 0.002 = 0 [/tex]

    And rearranged to get the same answer as my previous q' after subbing in my original q. (multiplying both sides by q.)
     
  5. Oct 26, 2010 #4

    Mark44

    Staff: Mentor

    You have a sign error.
    From your last equation in the previous post,
    q'/q = 3/p + 0.002 ==> q' = q(3/p + 0.002)
     
  6. Oct 26, 2010 #5
    According to [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/80/fa9842dd3479ee1baaab655e02a1c41.png [Broken]

    Isnt it q'/q = 3/p - 0.002 ??

    Thanks for your assistance so far by the way.

    Edit: oh from the post I just made, yes i see that, but the answer is still wrong (online submission, unlimited tries) the last post is incorrect, the original post is it's correct form
     
    Last edited by a moderator: May 5, 2017
  7. Oct 26, 2010 #6

    Mark44

    Staff: Mentor

    I was going by what you had in post #3 which I didn't notice was different from post #1.

    What you have in post #2 should be simplified.
    [tex] E\left(p\right) = -1 * \left[\left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right)\right] * \frac{p}{e^\left(3ln(p) - 0.002p +7\right)} [/tex]
    Cancel the e3lnp - .002p + 7 factors in the numerator and denominator.
     
  8. Oct 26, 2010 #7
    My god, that's all I had to do, after that I had the correct answer.

    Thank you for your help mark :) Must be tired or something.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook