Question

A factory installs new equipment that is expected to generate savings at the rate of 800e−0.2t dollars per year, where t is the number of years that the equipment has been in operation.(a) Find a formula for the total savings that the equipment will generate during its first t years.S(t) = _______________________(b) If the equipment originally cost $1000, when will it "pay for itself"? (Round your answer to one decimal place.)t =__________________ yr

Answer 1

a)
`S(t)=4000(1 - e^(-0.2t))`

b)
`t \approx 1.4 \text{ years}`

Explanation:

We are given the following information in the question:

`\displaystyle(dS(t))/(dt) = 800e^(-0.2t)`

a) We have to find formula for the total savings that the equipment.

`\displaystyle(dS(t))/(dt) = 800e^(-0.2t)\\\\dS(t) = 800e^(-0.2t)~dt\\\text{Integrating both sides}\\\int dS(t) = \int^t_0 800e^(-0.2t)~dt\\\\S(t) = \Bigg[800(e^(-0.2t))/(-0.2)\Bigg]^t_0 = [-4000e^(-0.2t)]^t_0 \\\\S(t) = -4000e^(-0.2t) + 4000 = 4000(1 - e^(-0.2t))\\S(t) = 4000(1 - e^(-0.2t))`

b) We have to equate savings to 1000$, so that it could pay for itself.

`S(t) =1000 =4000(1 - e^(-0.2t))\\\\(1 - e^(-0.2t)) = \displaystyle(1)/(4)\\\\e^(-0.2t) = (3)/(4)\\\\t = -(1)/(0.2)\ln (3)/(4) = 5 \ln (4)/(3) \approx 1.4`