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Math Help - Revenue problem

  1. #1
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    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?
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  2. #2
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    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?
    ...
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  3. #3
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    I'm not sure where you got p=-\frac{b}{2a}, but you are correct that we start by finding the first derivative of R. As R is differentiable, its maximum will occur either at a boundary point or a point at which

    R'=180q-3q^2=0.

    Disregarding the solution q=0 and dividing, we obtain

    180-3q=0

    for the nonzero solution.

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

    \frac{d}{dq}R'=\frac{d}{dq}(180q-3q^2)=180-6q=0.
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