Question 1118184: You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was 570 log-ons per month. When you lowered the price to $2.50, the demand increased to 855 log-ons per month.
(a) Construct a linear demand function for your Web site 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 $40 to maintain your site. Express your monthly profit P as a function of the log-on fee x. HINT [See Example 4.]
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?
$
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Demand graph has two points (3, 570) and (2.5, 855)
This has a slope of 285/-0.5 or -570
using point slope formula y-y1=m(x-x1), m slope, (x1, y1) a point
y-570=-570(x-3)
y-570=-570x+1710
R(x)=-570x+2280
Profit function is Revenue-cost,
Revenue
3--------570
2.5------855
2-------1140
Let x be the amount different from 3.
This is (3-x)(570)(1+x)-40
That is 570(3+2x-x^2)-40
-570x^2+1140x+1670
the x-value is -b/2a at maximum or -1140/-1140 or x=$1 different from 3. This will produce greatest at $2, where x=1, and P(x)=2(570)(2)-40=$2240.
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