Math Problem Statement

You have $99,000 you would like to invest in three different stocks: MarkyB Inc., JohnJohn Ltd., and Garretts Spaghetti House. You would like to invest no more than $11,000 in Garretts Spaghetti House, and you want the amount you invest in MarkyB Inc., to be at least the amount invested in JohnJohn Ltd. and Garretts Spaghetti House combined. If MarkyB Inc. earns 16% annual interest, JohnJohn Ltd. earns 15% annual interest, and Garretts Spaghetti House earns 9% 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 earned?

Solution

We can solve this investment problem using the method of linear programming to maximize your annual interest earned. Let's define the variables for the amounts you will invest in each of the three stocks:

  • Let xx be the amount invested in MarkyB Inc..
  • Let yy be the amount invested in JohnJohn Ltd..
  • Let zz be the amount invested in Garretts Spaghetti House.

Objective:

The goal is to maximize the annual interest earned, which can be written as the function: Maximize: P=0.16x+0.15y+0.09z\text{Maximize: } P = 0.16x + 0.15y + 0.09z where PP is the total annual interest earned.

Constraints:

  1. Total investment constraint: x+y+z=99,000x + y + z = 99,000
  2. Investment limit in Garretts Spaghetti House: z11,000z \leq 11,000
  3. Investment constraint for MarkyB Inc. (the amount invested in MarkyB Inc. should be at least the combined investment in JohnJohn Ltd. and Garretts Spaghetti House): xy+zx \geq y + z
  4. Non-negativity constraints (since you cannot invest a negative amount): x0,y0,z0x \geq 0, \quad y \geq 0, \quad z \geq 0

Solving the Problem:

We will now solve the system of constraints to find the optimal values of xx, yy, and zz that maximize the interest.

Step 1: Substituting the total investment constraint

From the equation x+y+z=99,000x + y + z = 99,000, we can express xx as: x=99,000yzx = 99,000 - y - z This allows us to substitute into other constraints.

Step 2: Substituting into the MarkyB Inc. constraint

Using xy+zx \geq y + z and substituting x=99,000yzx = 99,000 - y - z, we get: 99,000yzy+z99,000 - y - z \geq y + z Simplifying: 99,0002y+2z99,000 \geq 2y + 2z 49,500y+z49,500 \geq y + z Thus, we have another constraint: y+z49,500y + z \leq 49,500

Step 3: Analyzing Garretts Spaghetti House constraint

We know that z11,000z \leq 11,000. Combining this with y+z49,500y + z \leq 49,500, we can deduce the maximum allowable yy.

Now, let's proceed to solve this system to maximize the objective function P=0.16x+0.15y+0.09zP = 0.16x + 0.15y + 0.09z.

Let me calculate the optimal values for xx, yy, and zz.The optimal investment strategy is to:

  • Invest $99,000 in MarkyB Inc..
  • Invest $0 in JohnJohn Ltd..
  • Invest $0 in Garretts Spaghetti House.

This strategy will yield the maximum annual interest of $15,840.

It appears that the best option is to invest entirely in MarkyB Inc., as it offers the highest interest rate at 16%.

Would you like more details on the solution, or do you have any questions?

Here are 5 related questions for further exploration:

  1. How would the solution change if Garretts Spaghetti House had a higher interest rate?
  2. What if there were a minimum investment requirement for JohnJohn Ltd.?
  3. How would the investment strategy change if you were required to diversify among all three stocks?
  4. What is the impact of reducing the total amount available for investment?
  5. How would taxes or fees affect the final interest earned?

Tip: In linear programming, maximizing returns often leads to investing heavily in the option with the highest rate, provided constraints allow it.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Algebraic Constraints

Formulas

Objective function: P = 0.16x + 0.15y + 0.09z
Total investment constraint: x + y + z = 99,000
MarkyB Inc. constraint: x ≥ y + z
Garretts Spaghetti House constraint: z ≤ 11,000

Theorems

Linear Programming Maximization
Non-negativity constraints

Suitable Grade Level

Grades 11-12