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? 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 out the constraints

We have the following constraints: Total investment: $x + y + z = 99000$ Garretts Spaghetti House maximum: $z \le 16000$ MarkyB Inc. minimum: $x \ge y + z$ All investments must be non-negative: $x, y, z \ge 0$

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

4. Analyze the constraints to find the optimal strategy

Since MarkyB Inc. has the highest interest rate, we want to maximize the investment in it, while still meeting all constraints. We see $x \ge y + z$. Replacing $x$ in our first equation we get $x+y+z = 99000 \implies y+z+y+z\le99000 \implies 2y+2z \le 99000 \implies y+z \le 49500$.

We know that the constraint $z \le 16000$. Also, we can deduce that $x=99000 - y - z$.

5. Consider the edge cases

To begin, we know that $x=y+z$, so we will have no additional money being put into JohnJohn Ltd and Garretts Spaghetti House unless we can meet the other conditions.

Our equation becomes $x+y+z = 99000$, and we know that $I = 0.17x+0.15y+0.08z$. We also can write the equation to equal $0.17(y+z) + 0.15y+0.08z=0.32y+0.25z$. To make $I$ the largest, we need $y$ to be the biggest number and $z$ to be the biggest number. $z$ can be at most $16000$, so let's make it that. Thus, $0.32y+0.25(16000) = 0.32y+4000$. Since $y+z=49500$ and since we know $z=16000$, we know $y=33500$. $x=y+z=33500+16000 = 49500$ $x+y+z = 49500+33500+16000 = 99000$ $I=0.17(49500) + 0.15(33500)+0.08(16000) = 8415+5025+1280 = 14720$.

Now we can see if we get a higher interest if we put $z=0$. If $z=0$, we have the equations $x \ge y$ and $x+y=99000$. Thus $x \ge 99000-x$ implying $x \ge 49500$, but we are only looking for $x$ to be the smallest amount, thus $x = 49500$ and $y=49500$. $I = 0.17(49500)+0.15(49500) = 8415+7425=15840$

Thus the answer is when $z=0$ and x=y=49500, so $I = 15840$.