Another maximize

  1. tony873004

    tony873004 1,555
    Science Advisor
    Gold Member

    The demand for rubies is given by the equation
    [tex]
    q = \frac{4}{3}p + 80
    [/tex]

    where p is the price and q is the number of rubies sold each week. At what price should the rubies be sold to maximize weekly revenue?

    [tex]
    \begin{array}{l}
    R = pq \\
    R = p\left( {\frac{4}{3}p + 80} \right) \\
    R = \frac{4}{3}p^2 + 80p \\
    \\
    R' = \frac{8}{3}p + 80 \\
    \\
    \frac{8}{3}p + 80 = 0 \\
    \\
    \frac{8}{3}p = - 80 \\
    \\
    p = \frac{{ - 80}}{{\left( {\frac{8}{3}} \right)}} = \frac{{ - 240}}{8} = - 30 \\
    \end{array}
    [/tex]
    To maximize weekly revenue, they should give away the rubies and $30 per rubie. (Obviously wrong. The back of the book says $30)
     
  2. jcsd
  3. It looks like by the second derivative of R with respect to P that actually minimizes the revenue. To maximize revenue then would be to choose the highest price possible. The demand formula doesn't make sense to begin with because it implies that you sell more rubies as the price goes up on rubies, so a strange answer doesn't seem out of the question.
     
    Last edited by a moderator: May 20, 2006
  4. This doesn't make any sense.

    q = 4/3p + 80

    So when you increase the price you sell more rubies? In that case you would maximize weekly revenue by selling an infinite number of rubies. Either I'm not understanding you correctly, or you are missing a minus sign in that formula, try something like:

    q = -4/3p + 80

    and see how that works.

    EDIT: That seems to give you the right answer, it must be a typo in the book.

    ~Lyuokdea
     
    Last edited: May 20, 2006
  5. dav2008

    dav2008 624
    Gold Member

    To go along with what Lyokdea said, you also know something is wrong when you look at your equation for revenue that you're trying to maximize: R=4/3p2+80p. This function clearly has no absolute maximum; only a minimum. (It's an upwards-opening parabola)
     
  6. nrqed

    nrqed 3,048
    Science Advisor
    Homework Helper

    :rofl: :rofl: You made me laugh out loud with your last sentence :biggrin:

    Are you sure that the first equation should not read -4/3p+80?
    With the equation you gave, increasing the price keeps increasing the number sold, which is weird. And notice that your function for R is concave upward..Taking the price to infinity will lead to infinite revenues!
    (what you found is actually the minimum).

    If the equation for R was -4/3p + 80 it would fix everything and make sense now. Maybe a typo in the book?

    Patrick
     
    Last edited: May 20, 2006
  7. tony873004

    tony873004 1,555
    Science Advisor
    Gold Member

    sorry, I was blind and didn't see the negative sign. -4/3p + 80. Sorry for the inconvenience.
     
  8. nrged, you mean -4/3p + 80, not -3/4p right? That's what came out to match his answer when I ran through the math.

    ~Lyuokdea
     
  9. tony873004

    tony873004 1,555
    Science Advisor
    Gold Member

    I really need to learn to spot stuff like this rather than just diving straight into the formulas.
     
  10. nrqed

    nrqed 3,048
    Science Advisor
    Homework Helper

    Yes, absolutely. I will correct it. That was a typo. Thanks for letting me know.
    And sorry for having repeated what you said, I just saw your post after I sent mine.

    Regards
     
  11. Curious3141

    Curious3141 2,970
    Homework Helper

    If you were a jeweller giving away free rubies plus $30, I'll take whatever you got. :rofl:
     
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