1. ## recursive formula help

Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

Any help?

2. Originally Posted by iknowyou
Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

Any help?
The question is a little unclear, I think by "maintains" they mean "starts out with." Also it's unclear whether the new plants are included when calculating that 10% get sold. I'll assume not. Additionally, it's not clear whether 50,000 should be the number of plants for year zero or for year one! I'll assume one.

So we can say

f(1) = 50000

for all integers n > 1: f(n) = 0.9 * f(n-1) + 4000

I get f(20) approximately equals 41350.85

3. Originally Posted by iknowyou
Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

Any help?
Sorry- for some reason I was thinking "sold" rather than "planted" 4000 new plants. Where I had "-4000" I have changed to "+4000".

Just write down exactly what is said. If in year n, the company maintains $A_n$ plants, sells 10% of them, $0.1A_n$, and [b]plants 4000, then in year n+1, the company will have $A_{n+1}= A_n- 0.1A_n+ 4000= 0.9A_n+ 4000$. As undefined says, the "50,000 plants" must be some initial value. It wouldn't make sense to say the maintain 50,000 plants every year and then ask how many they maintain in the twentieth year!

Given that $A_0= 50000$ and that $A_{n+1}= 0.9A_n+ 4000$, can you find $A_{20}$?

4. Originally Posted by HallsofIvy
Just write down exactly what is said. If in year n, the company maintains $A_n$ plants, sells 10% of them, $0.1A_n$, and sells 4000, then in year n+1, the company will have $A_{n+1}= A_n- 0.1A_n- 4000= 0.9A_n- 4000$. As undefined says, the "50,000 plants" must be some initial value. It wouldn't make sense to say the maintain 50,000 plants every year and then ask how many they maintain in the twentieth year!

Given that $A_0= 50000$ and that $A_{n+1}= 0.9A_n- 4000$, can you find $A_{20}$?
Why did you subtract 4000 instead of add? Was that a typo or did I misinterpret what it means to plant new plants?

5. Hello, iknowyou!

undefined is right . . . The wording is not clear.
I'll make my own interpretation and solve it.

A company starts with 50,000 plants and sells 10% annually.
They plant 4000 new plants each year.
(a) Write a recursive formula for the number of plants in year $\,n$
(b) Find the number of plants being maintained in the 20th year.

Each year they have 90% of the previous year's plants ... plus 4000 new plants.

(a) We have: . $\boxed{f(n) \;=\;\tfrac{9}{10}f(n\!-\11) + 4000}$ .[1]

Write the equation for the "next year":
. . . . . . . . . . $f(n\!+\!1) \;=\;\frac{9}{10}f(n) + 4000$ .[2]

Subtract [2] - [1]: . $f(n\!+\!1)-f(n) \;=\;\frac{9}{10}f(n) - \frac{9}{10}f(n\!-\!1)$

and we have: . $f(n\!+\!1) - \frac{19}{10}f(n) + \frac{9}{10}f(n\!-\!1) \;=\;0$

Let $f(n) \,=\,X^n$

We have: . $X^{n+1} - \frac{19}{10}X^n + \frac{9}{10}X^{n-1} \;=\;0$

Divide by $X^{n-1}\!:\;\;X^2 - \frac{19}{10}X + \frac{9}{10} \;=\;0$

Factor: . $(X - 1)\left(X - \frac{9}{10}\right) \;=\;0$

Hence: . $X \,=\,1,\;X \,=\,\frac{9}{10}$

The generating function is of the form: . $f(n) \;=\;A + B\left(\frac{9}{10}\right)^n$

We know that: . $f(0) = 50,\!000,\;\;f(1) = 49,\!000$

We have: . $\begin{Bmatrix}f(0) = 50,\!000\!: & A + B &=& 50,\1000 & [3] \\ f(1) = 49,\!000\!: & A + \frac{9}{10}B &=& 49,\!000 & [4] \end{Bmatrix}$

Subtract [3] - [4]: . $\frac{1}{10}B \:=\:1,\!000 \quad\Rightarrow\quad B \:=\:10,\!000$

Substitute into [3]: . $A + 10,\!000 \:=\:50,\!000 \quad\Rightarrow\quad A \:=\:40,\!000$

The generating function is: . $f(n) \;=\;40,\!000 + 10,\!000\left(\frac{9}{10}\right)^n$

(b) In the 20th year: . $f(20) \;=\;40,\1000 + 10,\!000\left(\frac{9}{10}\right)^{20} \;\approx\;\boxed{41,216}$

6. Yes, that was a typo. I have changed it to +4000.