Question

the monthly income I, in dollars, from a new product is given by

I(t) = 8900 - 6500e^-0.007t

^{where t is the time, in months, since the product was first put on the market.}

^{(a) What was the monthly income after the 10th month and after the 100th month? Round to nearest cent.}

^{(b) What will the monthly income from the product appraoch as the time increases without bound?}

Answer 1

Given the monthly income value is as a function of months

I(t) = 8900 - 6500e^-0.007t

a) We have to find the monthly income after the 10 months i.e t =10 and the monthly income after 100th month

i.e t =100

at t = 10,

I(10) = 8900 - 6500e^-0.007(10)

=8900-6500(0.9323938199)

8900-6060.5598

I(10) = $2839.44

I(100) after 100 months is,

I(100) = 8900 - 6500e^-0.007(100)

= 8900 - 6500e^-0.7

=5672.1955

I(100) = $5672.20

b) As we see, the months are getting increased, the value of 6500e^-0.007t is approaching to zero.

So, the income approaches to 8900 as time increases without bound