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?
$

(a) To find the monthly income after the 30th month, we can substitute t = 30 into the equation I(t) = 81000 − 45000e^(-0.005t):

I(30) = 81000 − 45000e^(-0.005 * 30)
≈ 81000 − 45000e^(-0.15)
≈ 81000 − 45000(0.860708653)
≈ 81000 − 38731.8879
≈ $42,268

Therefore, the monthly income after the 30th month is approximately $42,268.

To find the monthly income after the 130th month, we can substitute t = 130 into the equation I(t) = 81000 − 45000e^(-0.005t):

I(130) = 81000 − 45000e^(-0.005 * 130)
≈ 81000 − 45000e^(-0.65)
≈ 81000 − 45000(0.522995514)
≈ 81000 − 23534.7991
≈ $57,465

Therefore, the monthly income after the 130th month is approximately $57,465.

(b) To determine what the monthly income from the product approaches as the time increases without bound, we can inspect the equation I(t) = 81000 − 45000e^(-0.005t).

As t approaches infinity, the exponential term e^(-0.005t) approaches zero since the exponent becomes increasingly negative. Therefore, the income from the product approaches the constant term of 81000.

Hence, the monthly income from the product approaches $81,000 as time increases without bound.

To find the monthly income after a certain number of months, you need to substitute the value of t into the given equation, I(t) = 81000 - 45000e^(-0.005t). Let's calculate the monthly income after the 30th month and the 130th month.

(a) Monthly income after the 30th month (t = 30):
I(30) = 81000 - 45000e^(-0.005*30)
To calculate this, first evaluate -0.005*30, which gives you -0.15. Then, calculate e^(-0.15) using a scientific calculator or a calculator with an exponential function. Finally, multiply this value by 45000 and subtract it from 81000:

I(30) ≈ 81000 - 45000 * e^(-0.15)
I(30) ≈ 81000 - 45000 * 0.8607
I(30) ≈ 81000 - 38731.5
I(30) ≈ $42268.5 (rounded to the nearest dollar amount)

Therefore, the monthly income after the 30th month is approximately $42,268.

Monthly income after the 130th month (t = 130):
I(130) = 81000 - 45000e^(-0.005*130)
Similarly, evaluate -0.005*130, which gives -0.65. Then calculate e^(-0.65) and multiply that value by 45000. Finally, subtract the result from 81000:

I(130) ≈ 81000 - 45000 * e^(-0.65)
I(130) ≈ 81000 - 45000 * 0.5197
I(130) ≈ 81000 - 23386.5
I(130) ≈ $57613.5 (rounded to the nearest dollar amount)

Therefore, the monthly income after the 130th month is approximately $57,614.

(b) To determine the monthly income from the product as the time increases without bound, we need to find the limit of I(t) as t approaches infinity.

As t approaches infinity, the exponential term e^(-0.005t) will approach zero, since the base of the exponent is less than 1. Therefore, the monthly income will approach the constant term, which is 81000 dollars.

Thus, as time increases without bound, the monthly income from the product will approach $81,000.