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.

$R(x)=?$

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.

$P(x)=?$

Determine the log-on fee you should charge to obtain the largest possible monthly profit.

$x=\$?$

What is the largest possible monthly profit?

$\$?$

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.

$R(x)=?$

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.

$P(x)=?$

Determine the log-on fee you should charge to obtain the largest possible monthly profit.

$x=\$?$

What is the largest possible monthly profit?

$\$?$

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, $4=(-0.00185)540+b$

$4=-0.99+b\Rightarrow b=4.99$

${D}_{1}=-0.00185q+4.99\phantom{\rule{0ex}{0ex}}i.e.,R(x)=-0.00185x+4.99$ 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 $R(x)=\$10$, we have to find x.

$10=-0.00185x+4.99\Rightarrow x=\$2708$

monthly profit P as function of log-on fee x.

is $P(x)=\frac{4.99-R(x)}{0.00185}=2697.29-540.54R(x)$

$\therefore P(x)=2697.29-540.54R(x)$

If $R(x)=0$, largest possible monthly profit P(x) will be maximum.

$\text{largest}\text{}P(x)=\$2697.29$

- 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, $4=(-0.00185)540+b$

$4=-0.99+b\Rightarrow b=4.99$

${D}_{1}=-0.00185q+4.99\phantom{\rule{0ex}{0ex}}i.e.,R(x)=-0.00185x+4.99$ 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 $R(x)=\$10$, we have to find x.

$10=-0.00185x+4.99\Rightarrow x=\$2708$

monthly profit P as function of log-on fee x.

is $P(x)=\frac{4.99-R(x)}{0.00185}=2697.29-540.54R(x)$

$\therefore P(x)=2697.29-540.54R(x)$

If $R(x)=0$, largest possible monthly profit P(x) will be maximum.

$\text{largest}\text{}P(x)=\$2697.29$

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