Linear Programming

You have $10,000 to invest, and three different funds to choose from. The municipal bond fund has a 5% return, the local bank's CDs have a 7% return, and a high-risk account has an expected 10% return. To minimize risk, you decide not to invest any more than $1,000 in the high-risk account. For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs. What's the best way to distribute your money given these constraints?

Let's define our variables:

\(\begin{align*}x\end{align*}\) is the amount of money invested in the municipal bond at 5% return

\(\begin{align*}y\end{align*}\) is the amount of money invested in the bank's CD at 7% return

\(\begin{align*}10000 - x - y\end{align*}\) is the amount of money invested in the high-risk account at 10% return

\(\begin{align*}z\end{align*}\) is the total interest returned from all the investments, so \(\begin{align*}z = .05x + .07y + .1(10000 - x - y)\end{align*}\) or \(\begin{align*}z = 1000 - 0.05x - 0.03y\end{align*}\). This is the amount that we are trying to maximize. Our goal is to find the values of \(\begin{align*}x\end{align*}\) and \(\begin{align*}y\end{align*}\) that maximizes the value of \(\begin{align*}z\end{align*}\).

Now, let's write inequalities for the constraints:

You decide not to invest more than $1000 in the high-risk account - that means:

\(\begin{align*}10000 - x - y \le 1000\end{align*}\)

You need to invest at least three times as much in the municipal bonds as in the bank CDs - that means:

\(\begin{align*}3y \le x\end{align*}\)

Also, you can't invest less than zero dollars in each account, so:

\(\begin{align*}x & \ge 0\\ y & \ge 0\\ 10000 - x - y & \ge 0\end{align*}\)

To summarize, we must maximize the expression \(\begin{align*}z = 1000 - .05x - .03y\end{align*}\) using the constraints:

\(\begin{align*}& 10000 - x - y \le 1000 && && y \ge 9000 - x\\ & 3y \le x && && y \le \frac{x}{3}\\ & x \ge 0 && \text{Or in slope-intercept form:} && x \ge 0\\ & y \ge 0 && && y \ge 0\\ & 10000 - x - y \ge 0 && && y \le 10000 - x\end{align*}\)

Step 1: Find the solution region to the set of inequalities by graphing each line and shading appropriately. The following figure shows the overlapping region:

The purple region is the feasibility region where all the possible solutions can occur.

Step 2: Next we need to find the corner points of the feasibility region. Notice that there are four corners. To find their coordinates, we must pair up the relevant equations and solve each resulting system.

System 1:

\(\begin{align*}y = \frac{x}{3}\!\\ y = 10000 - x\end{align*}\)

Substitute the first equation into the second equation:

\(\begin{align*}\frac{x}{3} &= 10000 - x \Rightarrow x = 30000 - 3x \Rightarrow 4x = 30000 \Rightarrow x = 7500\\ y &= \frac{x}{3} \Rightarrow y = \frac{7500}{3} \Rightarrow y = 2500\end{align*}\)

The intersection point is (7500, 2500).

System 2:

\(\begin{align*}y = \frac{x}{3}\!\\ y = 9000 - x\end{align*}\)

Substitute the first equation into the second equation:

\(\begin{align*}\frac{x}{3} &= 9000 - x \Rightarrow x = 27000 - 3x \Rightarrow 4x = 27000 \Rightarrow x = 6750\\ y &= \frac{x}{3} \Rightarrow y = \frac{6750}{3} \Rightarrow y = 2250\end{align*}\)

The intersection point is (6750, 2250).

System 3:

\(\begin{align*}y = 0\!\\ y = 10000 - x\end{align*}\)

The intersection point is (10000, 0).

System 4:

\(\begin{align*}y = 0\!\\ y = 9000 - x\end{align*}\)

The intersection point is (9000, 0).

Step 3: In order to find the maximum value for \(\begin{align*}z\end{align*}\), we need to plug all the intersection points into the equation for \(\begin{align*}z\end{align*}\) and find which one yields the largest number.

(7500, 2500): \(\begin{align*}z = 1000 - 0.05(7500) - 0.03(2500) = 550\end{align*}\)

(6750, 2250): \(\begin{align*}z = 1000 - 0.05(6750) - 0.03(2250) = 595\end{align*}\)

(10000, 0): \(\begin{align*}z = 1000 - 0.05(10000) - 0.03(0) = 500\end{align*}\)

(9000, 0): \(\begin{align*}z = 1000 - 0.05(9000) - 0.03(0) = 550\end{align*}\)

The maximum return on the investment of $595 occurs at the point (6750, 2250). This means that:

$6,750 is invested in the municipal bonds.

$2,250 is invested in the bank CDs.

$1,000 is invested in the high-risk account.