Question
Asked Jul 10, 2019

Assume a person invests $12000 at the end of each year for 30 years at an annual interest rate of r.  The amount of money after 30 years is (see figure)

Assume a person wants $1000000 in the account after 30 years. 

a.  Shwo the minimum value of r required to meet the person's investment needs satisfieds the following equation

1,000,000r - 12,000(1+r)^30 + 12,000 = 0

b. Apply Newton's method to solve the equation in part (a) to find the interest rate required to meet the person's investment goal.

12,000 ((1+n)30 - 1)
-
A =
r
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12,000 ((1+n)30 - 1) - A = r

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check_circleExpert Solution
Step 1

Part (a)

The future value amount should be equal to A = $ 1,000,000.

Please see the white board.

The minimum value of r required to meet the person's investment needs should satisfy the equation on the white board and hence it must be a solution of this equation.

12,000[1r)--1,000,000
A =
12,000[1r0 -1]= 1,000,000
1,000, 000r 12,000(10 +12,000 0
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12,000[1r)--1,000,000 A = 12,000[1r0 -1]= 1,000,000 1,000, 000r 12,000(10 +12,000 0

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Step 2

Part (b)

Before we get into solving the equation by Newton's method, let's understand the method a little bit better.

If xn is an approximate solution of F(x) and F’(xn) ≠ 0 then the next approximate solution xn+1 is given by the expression on the White Board.

F(x
х, — х,
F'(x
п
n+1
n+1
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F(x х, — х, F'(x п n+1 n+1

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Step 3

Let's select F(x) to be: F(x) = 1,000,000x – 12,000(1+x)30 + 12,000

Please note that it'sthe same function as derived in part (a). We have simply replaced "r" by "x".

Hence, F’(x) = 1,000,000 –...

F(x)
n+1
1,000,000x 12,000(1+ x)30 +12,000
1,000,000-360,000(1+ x)9
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F(x) n+1 1,000,000x 12,000(1+ x)30 +12,000 1,000,000-360,000(1+ x)9

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Tagged in

Math

Calculus

Derivative

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