Results 1 to 3 of 3

Thread: Revenue problem

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    92

    Question Revenue problem

    Revenue for sales of rubber baby-buggy bumpers is given by
    R(q)=90q^2-q^3 for 0<q<70 .

    a.what is the maximum revenue?
    b.at what quantity is revenue increasing the fastest?


    I tried finding R'(q)=180q-3q^2
    and then from here found out p=-b/2a so that price(p) is set in R'(p) to find the maximum revenue.

    I am not quite sure what part b wants.Am I doing part a correct?Any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    Quote Originally Posted by mathcalculushelp View Post
    Revenue for sales of rubber baby-buggy bumpers is given by
    R(q)=90q^2-q^3 for 0<q<70 .

    a.what is the maximum revenue?
    b.at what quantity is revenue increasing the fastest?


    I tried finding R'(q)=180q-3q^2
    and then from here found out p=-b/2a so that price(p) is set in R'(p) to find the maximum revenue.

    180q - 3q^2 = 0

    3q(60 - q) = 0

    q = 0 , q = 60 ... which value of q maximizes R ?

    once you determine that fact, find R(q) for that value.

    I am not quite sure what part b wants.Am I doing part a correct?Any help?

    the question is asking for the value of q for which R'(q) is a maximum ...

    R''(q) = 180 - 6q = 0

    q = 30 ... will this value of q yield a maximum for R'(q) ?

    how can you tell?
    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    I'm not sure where you got $\displaystyle p=-\frac{b}{2a}$, but you are correct that we start by finding the first derivative of $\displaystyle R$. As $\displaystyle R$ is differentiable, its maximum will occur either at a boundary point or a point at which

    $\displaystyle R'=180q-3q^2=0.$

    Disregarding the solution $\displaystyle q=0$ and dividing, we obtain

    $\displaystyle 180-3q=0$

    for the nonzero solution.

    For the second problem, we are asked to find the rate of maximum increase of $\displaystyle R$, which amounts to finding the maximum of $\displaystyle R'$ over the interval $\displaystyle [0,70]$. As $\displaystyle R'$ is differentiable, we are again left with boundary points and points at which

    $\displaystyle \frac{d}{dq}R'=\frac{d}{dq}(180q-3q^2)=180-6q=0.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Marginal Revenue Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 4th 2011, 01:13 PM
  2. revenue problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 11th 2010, 05:46 PM
  3. Replies: 1
    Last Post: Nov 24th 2009, 12:43 PM
  4. Replies: 1
    Last Post: Nov 18th 2009, 07:16 PM
  5. Maximum revenue problem.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 8th 2009, 04:23 AM

Search Tags


/mathhelpforum @mathhelpforum