Question

A trust account manager will invest $30,000 in three accounts that have yields of 8%, 7%, and 10%. The goal is to earn an interest of 9% on the total investment, assuming all $30,000 will be invested.

a. Suppose x,y,x represent the dollar amounts invested in each of the three accounts. Set up a system of linear equations that represents the given information

b. Suppose $16,000 is invested in the 10% account. How much should be invested in the other two accounts to achieve the goal? Describe/show your method.

c. What are the minimum and maximum amounts that could be invested in the 10% account that allows us to achieve the goal? Describe/show method

Answer 1

a. Suppose x,y,x represent the dollar amounts invested in each of the three accounts. Set up a system of linear equations that represents the given information

First equation: x + y + z = 30,000

0.08x + 0.07y + 0.10z = 30,000 x 0.09

Hence, second equation: 0.08x + 0.07y + 0.10z = 2,700

b. Suppose $16,000 is invested in the 10% account. How much should be invested in the other two accounts to achieve the goal? Describe/show your method.

z = 16,000

Hence, First equation now: x + y = 30,000 - 16,000 = 14,000

and 0.08x + 0.07y + 0.10 x 16,000 = 2,700

second equation now: 0.08x + 0.07y =  1,100

0.08 x First equation - second equation gives: 0.01y = 20

Hence, y = 20 / 0.01 = $ 2,000

and x = 14,000 - 2,000 = $ 12,000

c. What are the minimum and maximum amounts that could be invested in the 10% account that allows us to achieve the goal? Describe/show method

From part (a)

First equation: x + y + z = 30,000

Second equation: 0.08x + 0.07y + 0.10z = 2,700

In order to maximize z, amongst x and y, retain the variable that has minimum contribution to the return function. So, set x = 0

First equation: y + z = 30,000

Second equation: 0.07y + 0.10z = 2,700

Second equation - 0.07 x first equation gives: 0.03z = 600; hence z = 600/0.03 = 20,000

In order to minimize z, amongst x and y, retain the variable that has maximum contribution to the return function. So, set y = 0

First equation: x + z = 30,000

Second equation: 0.08x + 0.10z = 2,700

Second equation - 0.08 x first equation gives: 0.02z = 300; hence z = 300/0.02 = 15,000

Hence the minimum and maximum amounts that could be invested in the 10% account that allows us to achieve the goal are $ 15,000 and $ 20,000 respectively.