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Thread: Revenue Function - Integrals

  1. #1
    Junior Member
    Feb 2009

    Revenue Function - Integrals

    The marginal-revenue function for a manufacturer's product is of the form
    for constands a and b, where r is total revenue received when q units are produced and sold.

    Find the demand function, and express it in the form p = f(q)

    I'm not sure where to even begin, any help is appreciated....thank you!
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  2. #2
    Member Black's Avatar
    Nov 2009
    To find the demand function $\displaystyle p$, you must find $\displaystyle r$, then use $\displaystyle p=\frac{r}{q}$.

    To solve the integral

    $\displaystyle r(q)=a\int\frac{dq}{e^q+b}$,

    you use the substitution $\displaystyle u=e^q+b$. Then $\displaystyle du=e^q\,dq=(u-b)\,dq$,so

    $\displaystyle r(q)=a\int\frac{dq}{e^q+b}=a\int\frac{du}{u(u-b)}=\frac{a}{b}\int\left(\frac{1}{u-b}-\frac{1}{u}\right)\, du$ ...
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  3. #3
    Junior Member
    Nov 2009
    Evaluate $\displaystyle \int \frac{a}{e^q + b} dq$ to find revenue in terms of quantity sold. You know that if you sell nothing then you make no revenue, so use the fact that q = 0 implies r = 0 to solve for the constant of integration. Since $\displaystyle r = pq = f(q)q$, once you have r you can find f(q).
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  4. #4
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    $\displaystyle \int \frac{a}{e^q + b} \, dq

    $\displaystyle a \int \frac{1}{e^q + b} \cdot \frac{e^{-q}}{e^{-q}} \, dq

    $\displaystyle a \int \frac{e^{-q}}{1 + be^{-q}} \, dq$

    $\displaystyle u = 1 + be^{-q}$

    $\displaystyle du = -be^{-q} \, dq$

    $\displaystyle -\frac{a}{b} \int \frac{-be^{-q}}{1 + be^{-q}} \, dq$

    $\displaystyle -\frac{a}{b} \int \frac{du}{u}

    $\displaystyle -\frac{a}{b} \ln|u| + C$

    $\displaystyle -\frac{a}{b} \ln(1+be^{-q}) + C$
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