Revenue Function

**lil_cookie** · February 21st 2010, 07:12 AM

The marginal-revenue function for a manufacturer's product is of the form
$ \frac{dr}{dq}=\frac{a}{e^q+b} $
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!

**Black** · February 21st 2010, 08:17 AM

To find the demand function $p$ , you must find $r$ , then use $p=\frac{r}{q}$ .

To solve the integral

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

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

$r(q)=a\int\frac{dq}{e^q+b}=a\int\frac{du}{u(u-b)}=\frac{a}{b}\int\left(\frac{1}{u-b}-\frac{1}{u}\right)\, du$ ...

**nehme007** · February 21st 2010, 08:36 AM

Evaluate $\int \frac{a}{e^q + b} dq$ 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 $r = pq = f(q)q$ , once you have r you can find f(q).

**skeeter** · February 21st 2010, 11:16 AM

$\int \frac{a}{e^q + b} \, dq $

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

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

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

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

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

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

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

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

Thread: Revenue Function - Integrals

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