Thread: demand function

I need help starting this, I tried looking at examples but I can't get it...

The latest demand equation for you gaming website is given by q= -400x + 1200, where q is the number of users who log on per month and x is the log on fee you charge. Your IP provider charges you Site maintenance fee $20 per month and high volume access fee 0.5 per log on. Find the monthly cost as a function of the log on fee x. Find the monthly profit as a function of x. What is the largest possible monthly profit? What is the log on fee to obtain the largest possible profit?

Originally Posted by jnm07

The latest demand equation for you gaming website is given by q= -400x + 1200, where q is the number of users who log on per month and x is the log on fee you charge. Your IP provider charges you Site maintenance fee $20 per month and high volume access fee 0.5 per log on.

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

The total cost is the IP's fixed monthly charge, plus the IP's total access fee. The IP's total access fee is the product of the IP's per-logon fee and the number of user logons.

Originally Posted by jnm07

Find the monthly profit as a function of x. What is the largest possible monthly profit?

The profit is the difference between the income and the outgo. The income is the product of the number of users and the per-user fee that you charge. The total outgo is the cost you figured above.

Originally Posted by jnm07

What is the log on fee to obtain the largest possible profit?

Take the derivative of your profit function, and find the max point.

If you get stuck, please reply showing your work and reasoning so far, starting with the algebra for the first two parts.

Thank you!

the thing that confuses me though is, i looked at an example, but there was no "x", rather R=pq for revenue. should I just change the -400x+1200 to -400p +1200?