You have $38,000 you would like to invest in two different stocks, Stock Boll and Stock Coff. The buying limit on Stock Coff is $13,200 and you want to spend at least four times as much money on Stock Boll as Stock Coff.
If Stock Boll earns 8% 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)?

Solution

Let's define the variables:

- Let $ x $ represent the amount of money invested in Stock Coff.
- The remaining money $ 38,000 - x $ will be invested in Stock Boll.

### Conditions:
1. **Stock Coff Limit**: The maximum amount invested in Stock Coff is $ 13,200 $, so $ x \leq 13,200 $.
2. **Stock Boll Condition**: You want to invest at least four times as much in Stock Boll as in Stock Coff. This can be written as $ 38,000 - x \geq 4x $.
3. **Objective**: Maximize the annual interest earned.

#### Step 1: Expressing Conditions
From the second condition, we can write:
\[
38,000 - x \geq 4x
\]
Simplifying:
\[
38,000 \geq 5x
\]
\[
x \leq \frac{38,000}{5} = 7,600
\]
Thus, the investment in Stock Coff must satisfy $ x \leq 7,600 $, due to the restriction from Stock Boll.

Also, from the limit on Stock Coff, $ x \leq 13,200 $. But we already found $ x \leq 7,600 $, which is the more restrictive condition, so $ x \leq 7,600 $.

Thus, $ x $ must be between $ 0 $ and $ 7,600 $.

#### Step 2: Maximizing Interest
Now, let's calculate the total interest earned. The interest from Stock Coff (earning 4%) is $ 0.04x $, and the interest from Stock Boll (earning 8%) is $ 0.08(38,000 - x) $.

The total interest $ I $ is:
\[
I = 0.04x + 0.08(38,000 - x)
\]
Expanding:
\[
I = 0.04x + 0.08 \times 38,000 - 0.08x
\]
\[
I = 0.04x + 3,040 - 0.08x
\]
\[
I = 3,040 - 0.04x
\]
This is a decreasing function of $ x $, meaning that to maximize the interest, we should minimize $ x $.

#### Step 3: Conclusion
To maximize interest, the smallest value of $ x $ is $ 0 $. However, based on the constraint that you must invest at least four times more in Stock Boll, the minimum $ x $ that satisfies $ 38,000 - x = 4x $ is $ x = 7,600 $.

So, you should invest $ x = 7,600 $ in Stock Coff and $ 38,000 - 7,600 = 30,400 $ in Stock Boll.

#### Step 4: Maximum Annual Interest
Now let's calculate the maximum annual interest:
- Interest from Stock Coff: $ 0.04 \times 7,600 = 304 $
- Interest from Stock Boll: $ 0.08 \times 30,400 = 2,432 $

Thus, the total maximum interest is:
\[
304 + 2,432 = 2,736
\]

### Final Answer:
- Invest **\$7,600** in Stock Coff and **\$30,400** in Stock Boll.
- The maximum annual interest is **\$2,736**.

Would you like more details or have any questions?

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Here are 5 related questions to deepen understanding:
1. How would the investment strategy change if the Stock Coff limit were increased?
2. What if Stock Boll's interest rate changed to 10%?
3. Can you graph the total interest as a function of investment in Stock Coff?
4. How would the interest change if both stocks earned the same interest rate?
5. What is the minimum amount of interest possible if you only invest in Stock Boll?

**Tip**: Always ensure you fully understand the constraints in optimization problems, as they often determine the solution.

You have $38,000 you would like to invest in two different stocks, Stock Boll and Stock Coff. The buying limit on Stock Coff is $13,200 and you want to spend at least four times as much money on Stock Boll as Stock Coff. If Stock Boll earns 8% 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)?

This optimization problem explores how to maximize annual interest earnings by investing $38,000 between two stocks, Coff and Boll, subject to constraints. With Stock Coff limited to $13,200 and requiring a four-to-one investment ratio favoring Stock Boll, learn how to solve this linear maximization problem using interest rate calculations. Suitable for high school students in Grades 9-12.

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