Chapter 7.6, Problem 46E

To determine

To find: the optimal amount that should be invested in each type of investment and the optimal return..

Answer to Problem 46E

Optimal interest obtained is $27000.

Explanation of Solution

Given:

An investor has up to $450000 to invest in two types of investments. Type A pays6% annually and type B pays 10% annually. At least one half of the total portfolio is to be allocated to type A investments and at least one fourth of the portfolio is to be allocated to type B investment.

Calculation:

Consider the following objective function and their constraints. Let investment in A be considered as $x$and investment in B be considered as $y$ .

Max $f\left(x,y\right)=0.06x+0.01y$

  $\begin{array}{c}x+y\le 4250000\\ x\ge 225000\\ y\ge 112500\end{array}$

Fie the vertices using the above constraints.

At $x=112500$

  $\begin{array}{c}x+y=450000\\ y=337500\end{array}$

At $x=225000$

  $\begin{array}{c}x+y=450000\\ y=225000\end{array}$

And at the intersection of $x\ge 225000$and $y\ge 112500$we get $\left(225000,112500\right)$

Thus the vertices obtained are;

  $\begin{array}{l}\left(225000,112500\right)\\ \left(225000,225000\right)\\ \left(337500,112500\right)\end{array}$

Plot these constraints using the vertices on a graph to find the feasible region.

  

Check at all these points for the maximum value of the objective function $f\left(x,y\right)$ .

At $\left(22500,112500\right),$

  $\begin{array}{c}f\left(x,y\right)=0.06x+0.01y\\ =0.06\times 225000+0.01\times 112500\\ =14625\end{array}$

At $\left(225000,225000\right),$

  $\begin{array}{c}f\left(x,y\right)=0.06x+0.01y\\ =0.06\times 225000+0.01\times 225000\\ =27000\end{array}$

At $\left(3375000,112500\right)$

  $\begin{array}{c}f\left(x,y\right)=0.06x+0.01y\\ =0.06\times 337500+\mathrm{0..01}\times 112500\\ =21375\end{array}$

Thus the optimal investment should be in half of the total for both types of investments. Their optimal interest obtained is $27000.