Question

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

q = −500x + 1200

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:

Site maintenance fee: $10 per month
High-volume access fee: $0.70 per log-on
Find the monthly cost as a function of the log-on fee x.

C(x) =
_______



Find the monthly profit as a function of x.

P(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?
$ _______

Answer 1

Answer:

C(x) =  $850 - $350x

Monthly Profit =   -500x² + 1550x - 850

x = $1.55

the largest possible monthly profit  = $351.25

Step-by-step explanation:

Given:

Demand equation:

q = -500x + 1200

where, q is the number of users who log on per month

x is the log-on fee you charge

Site maintenance fee =  $10 per month

High-volume access fee =  $0.70 per log-on

Now,

C(x) = $10 + ( $0.70 × q )

or

C(x) = $10 + ( $0.70 × (-500x + 1200) )

or

C(x) = $10 + ( - $350x + 840 )

or

C(x) =  $850 - $350x

Total monthly income = qx

or

Total monthly income = ( -500x + 1200 ) × x

or

Total monthly income = -500x² + 1200x

now,

Profit = Income - cost

or

Monthly Profit =   -500x² + 1200x - ( 850 - 350x )

or

Monthly Profit, P(x) =   -500x² + 1550x - 850

For the largest possible monthly profit

$\frac{d\textup{P(x)}}{\textup{dx}}$ = 0

or

$\frac{d(-500x^2+1550x-850)}{\textup{dx}}$ =0

or

-2 × 500x + 1550 = 0

or

-1000x + 850 = 0

or

x = $1.55

Now,

the largest possible monthly profit will be at x = $1.55

substitute in the function of profit

P(1.55) =   ( -500× 1.55² ) + ( 1550 × 1.55 ) - 850

or

the largest possible monthly profit = - 1201.25 + 2402.5 - 850

or

the largest possible monthly profit  = $351.25