The monthly income I, in dollars, from a new product is given by
I(t) = 75000 − 55000e^−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 60th month and after the 140th month?
I(60) = __________$
I(140) =_________ $
(b) What will the monthly income from the product approach as the time increases without bound?
____________$
replace t with 60and find l(t)
replace l(t) with 140 and take the ln of both sides to solve for t.
To find the monthly income after a specific number of months, we can use the given function I(t) = 75000 - 55000e^(-0.003t).
(a) To find the monthly income after the 60th month (t = 60), we substitute t = 60 into the function:
I(60) = 75000 - 55000e^(-0.003 * 60)
Using a calculator, we can evaluate e^(-0.003 * 60) to be approximately 0.4506.
I(60) = 75000 - 55000 * 0.4506 ≈ 75000 - 24783 ≈ $50217
Therefore, the monthly income after the 60th month is approximately $50217.
Next, to find the monthly income after the 140th month (t = 140), we substitute t = 140 into the function:
I(140) = 75000 - 55000e^(-0.003 * 140)
Using a calculator, we can evaluate e^(-0.003 * 140) to be approximately 0.1022.
I(140) = 75000 - 55000 * 0.1022 ≈ 75000 - 5611 ≈ $69389
Therefore, the monthly income after the 140th month is approximately $69389.
(b) To find what the monthly income from the product approaches as time increases without bound, we take the limit as t approaches infinity:
lim(t→∞) I(t) = lim(t→∞) (75000 - 55000e^(-0.003t))
As t approaches infinity, the term e^(-0.003t) approaches 0. Therefore:
lim(t→∞) I(t) = 75000 - 55000 * 0 = 75000
Thus, the monthly income from the product approaches $75000 as time increases without bound.