Chapter 7.3, Problem 92E

Solution

To calculate: The amount invested by him at each rate if he receives $\$1780$ interest the first year.

The amount invested at 10% APY is  $9000\le z\le \$20000$ and the amount invested at 6% is  $z-9000$  and the amount invested at 8% is  $29000-2z$ .

Given Information:

Mateo invests $20,000 in three investments earning 6%APY, 8°/0 APY, and 10°/0 APY. He invests $9000 more in the 10% investment than in the 6% investment. How much does he have invested at each rate if he receives $1780 interest the first year?

Calculation:

Consider the given information,

Suppose that x is representing amount invested at 6% APY, and y is representing amount invested at 8% APY, and z is representing amount invested at 10% APY.

No, write the equation, as Mateo invested $20000.

  $x+y+z=20000$   ...... (1)

And, he invested $9000 more in 10% investment then in the 6% investment.

  $\begin{array}{c}z=x+9000\\ -x+z=9000\end{array}$   ...... (2)

And, the total interest in the first year is $1780.

  $0.06x+0.07y+0.10z=1780$   ...... (3)

Now, write the system of the matrix as $AX=B$ .

  $A=\left[\begin{array}{ccc}1& 1& 1\\ -1& 0& 1\\ 0.06& 0.08& 0.10\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],\text{ and }B=\left[\begin{array}{c}20000\\ 9000\\ 1780\end{array}\right]$

The equation can be solved as,

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ X=A^{-1}B\end{array}$

Use the augmented matrix instead and apply row elementary operations to write it in reduced row echelon form:

  $\begin{array}{l}\left[\begin{array}{cccc}1& 1& 1& 20000\\ -1& 0& 1& 9000\\ 0.06& 0.08& 0.10& 1780\end{array}\right]\underrightarrow{R_2+R_1}\left[\begin{array}{cccc}1& 1& 1& 20000\\ 0& 1& 2& 29000\\ 0.06& 0.08& 0.10& 1780\end{array}\right]\\ \\ \underrightarrow{0.06R_1-R_3}\left[\begin{array}{cccc}1& 1& 1& 20000\\ 0& 1& 2& 29000\\ 0& -0.02& -0.04& -580\end{array}\right]\underrightarrow{-50R_3}\left[\begin{array}{cccc}1& 1& 1& 20000\\ 0& 1& 2& 29000\\ 0& 1& 2& 2900\end{array}\right]\\ \\ \underrightarrow{R_3-R_2}\left[\begin{array}{cccc}1& 1& 1& 20000\\ 0& 1& 2& 29000\\ 0& 0& 0& 0\end{array}\right]\underrightarrow{R_1-R_2}\left[\begin{array}{cccc}1& 0& -1& -9000\\ 0& 1& 2& 29000\\ 0& 0& 0& 0\end{array}\right]\end{array}$

Write the obtained result in the simplified form of the equations.

  $\begin{array}{l}x-z=-9000\\ y+2z=29000\end{array}$

And,

  $\begin{array}{c}x=z-9000\\ y=29000-2z\end{array}$

The last equation is a true statement which tells us that there are infinitely many solutions since it can use any real value for  z  to obtain its corresponding  x  and  y  values. Solving for  x  and  y  in terms of  z , the solution of the system is:

  $\left(z-9000,29000-2z,z\right)$

This implies that as long as the amount invested at 10% APY is  $9000\le z\le \$20000$ and the amount invested at 6% is  $z-9000$  and the amount invested at 8% is  $29000-2z$ .

Hence, the above solution is the required solution.