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 $16,000 in Garr... 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 $16,000 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 17% annual interest, JohnJohn Ltd. earns 15% annual interest, and Garretts Spaghetti House earns 8% 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? You should invest $ into MarkyB Inc., $ into JohnJohn Ltd., and $ into Garretts Spaghetti House, which will give you give you maximum annual interest $. Read more

Steps to Solve

1. Define the variables Let $x$ be the amount invested in MarkyB Inc., $y$ be the amount invested in JohnJohn Ltd., and $z$ be the amount invested in Garretts Spaghetti House.

2. Write down the constraints We have the following constraints:

   • Total investment: $x + y + z = 99000$
   • Garretts Spaghetti House limit: $z \le 16000$
   • MarkyB Inc. investment condition: $x \ge y + z$
   • Non-negativity: $x \ge 0$, $y \ge 0$, $z \ge 0$

3. Define the objective function The objective is to maximize the annual interest earned, which is given by: $I = 0.17x + 0.15y + 0.08z$

4. Transform the constraints

   • From $x + y + z = 99000$, we have $y = 99000 - x - z$.
   • Substitute $y$ in the inequality $x \ge y + z$: $x \ge (99000 - x - z) + z$, which simplifies to $2x \ge 99000$, or $x \ge 49500$.

5. Consider the extreme cases to maximize interest Since MarkyB Inc. has the highest interest rate, we want to maximize the investment in it. Also, given the constraint on Garretts Spaghetti House, we want to analyze cases where $z = 16000$ and where $z < 16000$.

6. Case 1: $z = 16000$ If $z = 16000$, then $x + y = 99000 - 16000 = 83000$. Also, $x \ge y + z$ becomes $x \ge y + 16000$, so $y \le x - 16000$. Substitute $y = 83000 - x$ into $y \le x - 16000$: $83000 - x \le x - 16000$, which gives $2x \ge 99000$, or $x \ge 49500$. Since $x + y = 83000$, we have $y = 83000 - x$. The interest earned in this case is $I = 0.17x + 0.15(83000 - x) + 0.08(16000)$. To maximize this, we need to maximize $x$. The largest possible amount for $x$ occurs when $y$ is minimized. Since $x \ge y+z$ and $z=16000$, $x$ needs to be at least 49500. The constraints are $x+y=83000$ and $x \ge y+16000$. Let $x = y + 16000$. Then $2y + 16000 = 83000$, so $2y = 67000$ and $y = 33500$. Then $x = 33500 + 16000 = 49500$. Then $I = 0.17(49500) + 0.15(33500) + 0.08(16000) = 8415 + 5025 + 1280 = 14720$. Now check $x = 83000 - y$. Since $x \ge y + 16000$, $83000 - y \ge y + 16000$, so $67000 \ge 2y$ and $y \le 33500$.

Let $y = 0$. Then $x = 83000$, $z = 16000$, $I = 0.17(83000) + 0.15(0) + 0.08(16000) = 14110 + 1280 = 15390$. However $83000 \ge 0 + 16000$ and $x = 83000, y = 0, z =16000$ and the sum is 99000.

Case 2: $x+y+z = 99000$. If MarkyB is at least JohnJohn + Garrett, invest the rest in MarkyB Maximize x. Let $x = y + z$, then $2x = 99000$, when y = z. It also means $x+x = 99000$. $x = 99000/2 = $49500$ To maximize interest earned, we want to make y zero, then $x = z$, meaning $2x = 99000$, so x + y + z is false.

So $Garretts Spaghetti House = 0$. and let $z = 0.$ We now have: $x+y = 99000$ and $x \ge y$, implying $x\ge 49500$

7. Optimal Solution From the above analysis, it is clear that to maximize the earning we need to invest as much as possible in MarkyB and the condition of $x \ge y+z$ needs to meet, and $x+y+z = 99000$, $z=16000$

Then, $x=83000, y=0, z=16000$. Interest = $0.17 * 83000 + 0.15 * 0 + 0.08 * 16000 = 14110 + 1280 = 15390$