Question

A homeowner installs a solar heating system, which generates 20 savings at a rate of 300e t dollars per year, t years after installation. Find a formula for the total savings after t years. If the system cost 30000 dollars to install, when will the savings match the installation cost

Answer 1

"35.84 years" is the appropriate solution.

Explanation:

The given values are:

Rate of savings generation,

`s'(t)=300e^{(t)/(20) }`

System cost to install,

= $30000

Now,

The total savings will be:

⇒
`s(t)=\int\limits s'(t) \ dt`

On substituting the values, we get

⇒
`= \int\limits 300e^{(t)/(20) } \ dt`

⇒
`s(t)=\frac{300e^{(t)/(20) }}{(1)/(20) }+c`

When,

Savings = 0

t = 0, then

⇒
`0=6000e^o+c`

⇒
`c=-6000`

The formula will be:

⇒
`s(t)=6000[e^{(t)/(20) }-1]`

On putting the values in the above formula, we get

⇒
`30000=6000(e^{(t)/(20) }-1)`

`5=e^{(t)/(20)}-1`

`e^{(t)/(20) }=6`

On taking log, we get

⇒
`(t)/(20) =log \ 6`

`t=20 log 6`

`=35.84 \ years`