Answer:
a)[tex]S(t)=4000(1 - e^{-0.2t})[/tex]
b) [tex]t \approx 1.4 \text{ years}[/tex]
Step-by-step explanation:
We are given the following information in the question:
[tex]\displaystyle\frac{dS(t)}{dt} = 800e^{-0.2t}[/tex]
a) We have to find formula for the total savings that the equipment.
[tex]\displaystyle\frac{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\frac{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})[/tex]
b) We have to equate savings to 1000$, so that it could pay for itself.
[tex]S(t) =1000 =4000(1 - e^{-0.2t})\\\\(1 - e^{-0.2t}) = \displaystyle\frac{1}{4}\\\\e^{-0.2t} = \frac{3}{4}\\\\t = -\frac{1}{0.2}\ln \frac{3}{4} = 5 \ln \frac{4}{3} \approx 1.4[/tex]
