Chapter 2.7, Problem 17E

To determine

To find: the maximum amount that should be deposited into each account in order to earn more interest possible in one year; if not possible state whether the problem is infeasible, unbounded or has alternate optimal solutions.

Answer to Problem 17E

First bank: $4000

City bank: $7000

Explanation of Solution

Given information:

Let the amount deposited in first bank $x$ and amount deposited in first bankbe $y$ .

Total amount in both the banks = $11000

Maximum deposit at first bank = $7500

Minimum deposit at city bank = $1000

Maximum deposit at city bank = $7000

Simple interest offered at first bank = 6%

Simple interest offered at city bank = 6.5%

Calculation:

Let $P\left(x,y\right)$ be the maximum interest function. The maximum interest is as follows:

  $P\left(x,y\right)=0.06x+0.065y$ .

The inequalities derived from the given information are as follows:

  $\begin{array}{c}x\le 75000\\ 1000\le y\le 7000\\ x+y\le 11000\end{array}$

Use a graphing calculator to plot the inequalities as shown below.

  

The vertices are $\left(0,1000\right),\left(7500,1000\right)\text{,}\left(7500,3500\right)\text{,}\left(4000,7000\right)\text{ and }\left(0,7000\right)$ . Now test each vertex as shown below.

  $\begin{array}{c}P\left(0,1000\right)=0.06\left(0\right)+0.065\left(1000\right)\\ =\$65\\ P\left(7500,1000\right)=0.06\left(7500\right)+0.065\left(1000\right)\\ =\$515\\ P\left(7500,3500\right)=0.06\left(7500\right)+0.065\left(3500\right)\\ =\$677.50\\ P\left(4000,7000\right)=0.06\left(4000\right)+0.065\left(7000\right)\\ =\$695\\ P\left(0,7000\right)=0.06\left(0\right)+0.065\left(7000\right)\\ =\$455\end{array}$

Thus, the maximum amount that should be deposited into first bank and city bank account in order to earn more interest possible in one year is $ 4000 and $ 7000 respectively.