# The monthly interest rate. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4.7, Problem 33E
To determine

## To calculate: The monthly interest rate.

Expert Solution

The interest rate is

x0.0076286

### Explanation of Solution

Given information:

The annuity formula A=Ri[1(1+i)n] where A represents the present value of annuity, R represent the payment amount, i is the interest rate per time period and n is the number of payments that will be made.

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Formula: s=rθ where s is arc length, r is the radius of the circle and θ is the central angle.

Law of cosine c2=a2+b22abcosθwhere a , b , c are sides of triangle .

Calculation:

Consider the,

A =$18000 R =$375

Payments are to be made every month for 5 years.

This implies

1year=12months5year=12×5months

=60months

Therefore, n =60

=18000=375i[1(1+i)60] (i)

Re-placing i by x we get,

=18000=375x[1(1+x)60]=18000x=375[1(1+x)60] = 18000x 375 =1 (1+x) 60 =48x=11(1+x)60=48x=(1+x)601(1+x)60 (ii)

=48x (1+x) 60 = (1+x) 60 1 =48x (1+x) 60 (1+x) 60 +1=0 . (iii)

Hence, we get

g(x)=48 (1+x) 60 (1+x) 60 +1 g(x)=48 (1+x) 60 (1+x) 60 +1

Now, let initial approximation be x1=0.1

For n=1

x2=x1g(x1)g'(x1)

x2=x148(1+x1)60(1+x1)60+12880(1+x1)5960(1+x1)59

x2=(0.1)48(1+0.1)60(1+0.1)60+128801(1+0.1)5960(1+0.1)59x2=(0.1)48(1.1)60(1.1)60+128801(1.1)5960(1.1)59

The second approximation is x2=0.85101094 .

Following with the iterations after 14 iterations will get the approximate value of x .

Hence , the monthly interest to be pay is :-

x0.0076286

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