Helppp! Maximum Profit problem?

Question

A company begins an advertising campaign in a certain city to market a new product. The percentage of the target market that buys the product is a function of the length of the advertising campaign. The company estimates this percentage as 1-e^-0.05t where t=number of days of the campaign. The target market is estimated to be 1,000,000 people and the price per unit is $0.40. The cost of advertasing is $1000 per day.
Find the length of the advertising campaign that will result in the maximum profit.

P.S. Please list all the steps taken to answer the problem. Im interested in learning also, not just getting the answer.

Answer 1

Cut-off time for advertisement company is when revenue at that day is equal to cost. Before that day company still has a profit, after that day it start losing money. To find revenue at day t we have to find number of people willing to by a product: 1000000(1- exp(-0.05t)). Each person brings revenue $0.40, so total revenue at this day is 400000(1- exp(-0.05t)). Expenses always 1000$. So cut-off day condition can be written as
400000(1- exp(-0.05t)) = 1000
1- exp(-0.05t) = 1/400
exp(-0.05t) = 399/400
0.05t = ln(400/399).
Finally t = 20*ln(400/399).

Answer 2

Sales Revenue = # of people that buy * product price

= size of T.M * % that buy * product price

= 1 x 10^6 * (1 - e^-0.05t) * 0.4

= 4 x 10^5 * (1 - e^-0.05t)

Cost of advertising campaign = 1000t

Profit = Revenue - Cost

P(t) = 4 x 10^5 * (1 - e^-0.05t) - 1000t

In order to find the value of t that maximizes profit, take the first derivative of this equation and set it equal to 0 (...if you take a line and make it tangent to the function's maximum point, the slope of that line would be 0)

P'(t) = 400000*0.05*e^-0.05t - 1000 = 0

20000e^-0.05t - 1000 = 0

e^-0.05t = 1000/20000 = 0.05

To easily solve for t, take the natural log of each side.....

ln (e^-0.05t) = ln (0.05)

-0.05t = ln (0.05)

t = 59.9 days

The advertising campaign needs to be run for approximately 60 days in order to maximize profit from the new product