Download You operate a gaming Web site, www.mudbeast.net, where users

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on.
When you charged $4 the demand was 570 log-ons per month. When you lowered the price to
$3.50, 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.
Solution: To construct the linear demand function 𝑞 = 𝑚𝑥 + 𝑏, use the given information to find
the slope 𝑚:
𝑞2 − 𝑞1 855 − 570
285
𝑚=
=
=
= −570.
𝑥2 − 𝑥1
3.5 − 4
−0.5
Then, use the slope to find 𝑏: 𝑞 = −570𝑥 + 𝑏, and 570 = −570(4) + 𝑏, so 𝑏 = 2850. The
demand function is 𝑞 = −570𝑥 + 2850. Next, recall that Revenue = (price) × (quantity), so
𝑅 = 𝑥𝑞 = 𝑥(−570𝑥 + 2850) = −570𝑥 2 + 2850𝑥.
(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.
Solution: Recall that Profit = Revenue − Cost, and the only cost associated with this problem is
the $40 monthly fee. Thus,
𝑃 = −570𝑥 2 + 2850𝑥 − 40
Determine the log-on fee you should charge to obtain the largest possible monthly profit.
Solution: The profit function is quadratic, with leading coefficient that’s negative. Therefore, it
opens downward and its vertex is the maximum point. So, we must find the 𝑥-coordinate of the
vertex to find the log-on fee that will maximize the monthly profit. Use the vertex formula to do
this (found on p. 621 of the textbook):
−2850
𝑥max =
= 2.5
2(−570)
Thus, they should charge $2.50.
What is the largest possible monthly profit?
Solution: Here we need to find the 𝑃-coordinate of the vertex. Do this by plugging the 𝑥coordinate back into the profit function:
𝑃max = −570(2.5)2 + 2850(2.5) − 40 = 3522.5
Thus, the largest monthly profit is $3,522.50.