Results 1 to 2 of 2

Math Help - still cant get the answer....

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    16

    still cant get the answer....

    Hi I cant solve these two optimization problems...


    1. suppose that 20,000 fans will go to a ball game when the price of a ticket is $5, and that 500 fewer fans will go each $1 increase in price. What should the ticket price be in order to maximize revenue?



    2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.


    Can anyone solve these?

    Thanks
    - for the first one I set up in y-y1 = m(x-x1) form. with the points being (5,2000) and (6,19500). It came out to be y=-500x+22500. I then did R(x)=xp R(x)=x(-500x+22500) and distributed the x. Then i took the first derivitve and solved for 0. I got 22.5 but I think I solved for something else because that price is too high.
    - As far as the second one I tried just plugging the two formulas in the formula p(x) = R(x)-c(x), taking the first derivitive and then solving for 0, but Im lost on it and really dont know what to do.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    Quote Originally Posted by mike21 View Post
    Hi I cant solve these two optimization problems...


    1. suppose that 20,000 fans will go to a ball game when the price of a ticket is $5, and that 500 fewer fans will go each $1 increase in price. What should the ticket price be in order to maximize revenue?



    2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.


    Can anyone solve these?

    Thanks
    - for the first one I set up in y-y1 = m(x-x1) form. with the points being (5,2000) and (6,19500). It came out to be y=-500x+22500. I then did R(x)=xp R(x)=x(-500x+22500) and distributed the x. Then i took the first derivitve and solved for 0. I got 22.5 but I think I solved for something else because that price is too high.
    - As far as the second one I tried just plugging the two formulas in the formula p(x) = R(x)-c(x), taking the first derivitive and then solving for 0, but Im lost on it and really dont know what to do.

    2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.


    Revenue'(x) = 3x^2-6x+2
    Cost'(x) = 2x^2 - x - 2

    Set them equal, then simplify:

    x^2 - 5x + 4 = 0

    (x-1)(x-4) = 0

    Two production levels are candidates for maximum profit. Pick the one where the second derivative of profit is negative (concave down, local maximum, by the second derivative test).

    Good luck!


    P'(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: March 3rd 2013, 08:17 PM
  2. Converting my answer into the books answer
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 10th 2011, 03:06 PM
  3. Replies: 1
    Last Post: October 4th 2010, 05:46 PM
  4. Replies: 3
    Last Post: April 6th 2008, 04:18 PM

Search Tags


/mathhelpforum @mathhelpforum