Vorbeckenuc

2022-07-27

You operate a gaming website, where users must pay a small fee to log on. When you charged $4 the demand was 540 log-ons per month. When you lowered the price to$3.50, the demand increased to 810 log-ons per month.
a) Construct a linear a linear demand function for your website and hence obtain the monthly revenue R as a function of the log-on fee x.
$R\left(x\right)=?$
b) Your internet provider charges you a monthly fee of $10 to maintain your site. Express your monthly profit P as a function of the log-on fee x. $P\left(x\right)=?$ Determine the log-on fee you should charge to obtain the largest possible monthly profit. $x=?$ What is the largest possible monthly profit? $?$ ### Answer & Explanation Marisa Colon Beginner2022-07-28Added 18 answers Step 1 - when you charged$4, demand was 540 log-ons per month,
- when you charged $3.50, demand was 810 log-ons per month. a) Consider linear demand function, ${D}_{q}=mq+b$, where q is quantity. Here we have two ordered pairs (540,4) and (810,3.5) first, we calculate slope of demand function. $m=\frac{3.5-4}{810-540}=\frac{-0.5}{270}=-0.00185$ $\therefore {D}_{q}=-0.00185q+b$ for (540,4), we have, $4=\left(-0.00185\right)540+b$ $4=-0.99+b⇒b=4.99$ ${D}_{1}=-0.00185q+4.99\phantom{\rule{0ex}{0ex}}i.e.,R\left(x\right)=-0.00185x+4.99$ is monthly revenue R as function of log-on fee x. Step 2 b) If internet provides charges a monthly fee of$10, so, here $R\left(x\right)=10$, we have to find x.
$10=-0.00185x+4.99⇒x=2708$
monthly profit P as function of log-on fee x.
is $P\left(x\right)=\frac{4.99-R\left(x\right)}{0.00185}=2697.29-540.54R\left(x\right)$
$\therefore P\left(x\right)=2697.29-540.54R\left(x\right)$
If $R\left(x\right)=0$, largest possible monthly profit P(x) will be maximum.

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