Answer:
"35.84 years" is the appropriate solution.
Step-by-step explanation:
The given values are:
Rate of savings generation,
[tex]s'(t)=300e^{\frac{t}{20} }[/tex]
System cost to install,
= $30000
Now,
The total savings will be:
⇒ [tex]s(t)=\int\limits s'(t) \ dt[/tex]
On substituting the values, we get
⇒ [tex]= \int\limits 300e^{\frac{t}{20} } \ dt[/tex]
⇒ [tex]s(t)=\frac{300e^{\frac{t}{20} }}{\frac{1}{20} }+c[/tex]
When,
Savings = 0
t = 0, then
⇒ [tex]0=6000e^o+c[/tex]
⇒ [tex]c=-6000[/tex]
The formula will be:
⇒ [tex]s(t)=6000[e^{\frac{t}{20} }-1][/tex]
On putting the values in the above formula, we get
⇒ [tex]30000=6000(e^{\frac{t}{20} }-1)[/tex]
[tex]5=e^{\frac{t}{20}}-1[/tex]
[tex]e^{\frac{t}{20} }=6[/tex]
On taking log, we get
⇒ [tex]\frac{t}{20} =log \ 6[/tex]
[tex]t=20 log 6[/tex]
[tex]=35.84 \ years[/tex]
