The monthly income I, in dollars, from a new product is given by
I(t) = 81000 − 45000e^-0.005t 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 30th month and after the 130th month?
I(30) = $
I(130) = $
(b) What will the monthly income from the product approach as the time increases without bound?
$
I(30) = 81,000 - 45,000e^-0.005t =
81,000 - 45,000*e^(-0.15) = $42,268.14
I(130) = 81,000 - 45,000*e^(-0.65) =
b. I(approach): 81,000 - 0 = $81,000
To find the monthly income after the 30th month, we can substitute t = 30 into the equation I(t) = 81000 − 45000e^(-0.005t).
(a) Calculating the monthly income after the 30th month:
I(30) = 81000 - 45000e^(-0.005 * 30)
To find the monthly income after the 130th month, we can substitute t = 130 into the same equation.
I(130) = 81000 - 45000e^(-0.005 * 130)
To find the limit of the monthly income as time increases without bound (t approaches infinity), we need to take the limit as t approaches infinity of the equation I(t) = 81000 − 45000e^(-0.005t).
(b) Taking the limit as t approaches infinity:
lim(t->∞) [81000 - 45000e^(-0.005t)]
To evaluate these expressions, we can use a calculator or computer software with a numerical integration function.