Question 1081002: The monthly income I, in dollars, from a new product is given by
I(t) = 86000 − 65000e^−0.003t
where t is the time, in months, since the product was first put on the market. (Round your answers to the nearest dollar amount.)
(a) What was the monthly income after the 50th month and after the 100th month?
I(50) = $
I(100) = $
(b) What will the monthly income from the product approach as the time increases without bound?
$
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula is:
i(t) = 86000 - 65000 * e^-.003t
since e^-a = 1/e^a, this formula becomes:
i(t) = 86000 - 65000 / e^.003t
you can see that, as t gets extraordinarily large, the value of i(t) gets closer and closer to 86000, because the value of 65000 / e^.003t gets closer and closer to zero.
to find the income in the 50th and 100th month, just replace t with 50 and 100 respectively to get:
when t = 50, monthly income = 86000 - 65000 / e^.003*50 = 30053.98153
when t = 60, monthly income = 86000 - 65000 / e^.003*100 = 37846.81566
this can be shown graphically below:
as t approaches a higher and higher number, the value of y approaches 8600 as shown in the following graph:
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