YOu have $32,000 you would like to invest in two different stocks, Stock Boll and Stock Coff. The buying limit on Stock Coff is $12,600 and you want to spend at least four times as much money on Stock Boll as Stock Coff.
If Stock Boll earns 6% annual interest and Stock Coff earns 4% annual interest, how much money (in dollars) should you invest in each stock to maximize your annual interest earned? What is the maximum annual interest (in dollars)?
You will earn a maximum total annual interest of $ 

Let's define variables to represent the amounts of money invested in each stock:

- Let $ x $ be the amount invested in **Stock Coff**.
- Let $ y $ be the amount invested in **Stock Boll**.

We know the following:
1. The total investment is \$32,000, so:
   \[
   x + y = 32,000
   \]
2. The buying limit for Stock Coff is \$12,600, so:
   \[
   x \leq 12,600
   \]
3. You want to invest at least four times as much in Stock Boll as in Stock Coff, so:
   \[
   y \geq 4x
   \]
4. Stock Coff earns 4% annual interest and Stock Boll earns 6% annual interest. The total interest $ I $ earned is given by:
   \[
   I = 0.04x + 0.06y
   \]

### Step 1: Express $ y $ in terms of $ x $
From the first equation $ x + y = 32,000 $, we can express $ y $ as:
\[
y = 32,000 - x
\]

### Step 2: Substitute $ y $ into the inequality $ y \geq 4x $
Substituting $ y = 32,000 - x $ into $ y \geq 4x $ gives:
\[
32,000 - x \geq 4x
\]
Solving this inequality:
\[
32,000 \geq 5x
\]
\[
x \leq \frac{32,000}{5} = 6,400
\]

### Step 3: Determine the feasible range for $ x $
We have two constraints on $ x $:
- From the buying limit: $ x \leq 12,600 $
- From the inequality: $ x \leq 6,400 $

Thus, the maximum amount we can invest in Stock Coff is $ x = 6,400 $. The feasible range for $ x $ is:
\[
0 \leq x \leq 6,400
\]

### Step 4: Calculate the corresponding value of $ y $
When $ x = 6,400 $, we substitute this into $ y = 32,000 - x $ to get:
\[
y = 32,000 - 6,400 = 25,600
\]

### Step 5: Calculate the total interest earned
Now, we compute the total interest earned using $ x = 6,400 $ and $ y = 25,600 $:
\[
I = 0.04(6,400) + 0.06(25,600)
\]
\[
I = 256 + 1,536 = 1,792
\]

### Final Answer:
- **Amount to invest in Stock Coff**: \$6,400
- **Amount to invest in Stock Boll**: \$25,600
- **Maximum annual interest**: \$1,792

Do you need further details or clarifications? Here are five related questions to explore:

1. What happens to the total interest if the interest rates for both stocks are switched?
2. How would the solution change if you wanted to invest at least 3 times as much in Stock Boll instead of 4 times?
3. What is the effect on the maximum interest if the total amount to invest is increased to \$40,000?
4. How much less annual interest would you earn if the buying limit on Stock Coff were lowered to \$10,000?
5. What would be the total interest if Stock Boll's interest rate increased to 8%?

**Tip**: Always check your constraints carefully in optimization problems to ensure you're maximizing your return while staying within limits.

You have $32,000 you would like to invest in two different stocks, Stock Boll and Stock Coff. The buying limit on Stock Coff is $12,600 and you want to spend at least four times as much money on Stock Boll as Stock Coff. If Stock Boll earns 6% annual interest and Stock Coff earns 4% annual interest, how much money (in dollars) should you invest in each stock to maximize your annual interest earned? What is the maximum annual interest (in dollars)?

Discover how to maximize annual interest with an investment of $32,000 in two stocks, Stock Boll and Stock Coff. Learn how to apply linear inequalities and interest calculations to optimize your investment, considering constraints like a buying limit on Stock Coff and a requirement to invest at least four times as much in Stock Boll. Ideal for high school students in Grades 10-12.

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