The latest demand equation for your gaming website, www.mudbeast.net, is given b

Question

The latest demand equation for your gaming website, www.mudbeast.net, is given by

q = 300x + 1500

where q is the number of users who log on per month and x is the log-on fee you charge. Your Internet provider bills you as follows:

Find the monthly cost as a function of the log-on fee x.

Find the monthly profit as a function of x.

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

What is the largest possible monthly profit?
$

Explanation / Answer

The cost will be $40 plus 60 cents per logon.

so C = 40 + 0.6q
since we need Cost in terms of x,
we must plug in what we know about q in terms of x:
C = 40 + 0.5(-300x + 1500)
C = 40 - 150x + 750 = 790 - 150x

so our C(x) function is 790 - 150x

The monthly profit as a function of x will be:
the fee you charge times the # of logons - monthly cost.
In other words, P(x) = q*x - C(x)
so P(x) = -300x^2 + 1500x - (790 - 150x)
so P(x) = -300x^2 + 1650x - 790

To find the maximum profit, we take the derivative and set it equal to zero:
0 = -600x + 1650
x = -1650 / -600 = 2.75

so the maximum profit will come from charging $2.75 per log-on.

The maximum profit is P(2.75) = -300(2.75)^2 + 1650(2.75) – 790 = $1478.75.

Hope this helps.

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