Finding Present Value In Exercises 31-34, complete the table to determine the amount of money P that should be invested at rate r to produce a final balance of $100,000 in t years. See Example 5. n 1 2 4 12 365 Continuous compounding A r = 4.5 % , t = 30 years

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Calculus: An Applied Approach (Min...

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Ron Larson

Publisher: Cengage Learning

ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition

Ron Larson

Publisher: Cengage Learning

ISBN: 9781305860919

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Answer 5

To determine

To calculate: The amount A for corresponding value of n that is, number of times compounded per year in the provided table when $3000 is invested at the rate of 4.5% for 30 years:

Explanation of Solution

Given Information:

The amount of $3000 is invested at the rate of 4.5% for 30 years and the table is as follows:

n	1	2	4	12	365	Continuous compounding
A

Formula used:

The formula to compute the balance amount after t years when interest is compounded n times per year is,

A=P(1+rn)nt

Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals, t is the number of years and n is the number of times compounded per year.

The formula to compute the balance amount after t years when interest is compounded continuously is,

A=Pert

Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals and t is the number of years.

Calculation:

Consider that the amount of $3000 is invested at the rate of 4.5% for 30 years.

Here, P=$3000, r=4.5% and t=30 years.

Simplify the rate as,

r=4.5%=4.5100=0.045

When n=1,

Substitute 1 for number of times compounded per year, 0.045 for interest rate, $3000 for the amount deposited and 30 years for the time in the formula, “A=P(1+rn)nt” as,

A=P(1+rn)nt=$3000(1+0.0451)(1)(30)

Solve further as,

A=$3000(1.045)30≈$3000⋅3.745318≈$11,235.95

When n=2,

Substitute 2 for number of times compounded per year, 0.045 for interest rate, $3000 for the amount deposited and 30 years for the time in the formula, “A=P(1+rn)nt” as,

A=P(1+rn)nt=$3000(1+0.0452)(2)(30)

Solve further as,

A=$3000(1.0225)60≈$3000⋅3.800135≈$11,400.40

When n=4,

Substitute 4 for number of times compounded per year, 0.045 for interest rate, $3000 for the amount deposited and 30 years for the time in the formula, “A=P(1+rn)nt” as,

A=P(1+rn)nt=$3000(1+0

Finding Present Value In Exercises 31-34, complete the table to determine the amount of money P that should be invested at rate r to produce a final balance of $100,000 in t years. See Example 5. n 1 2 4 12 365 Continuous compounding A r = 4.5 % , t = 30 years

Calculus: An Applied Approach (Min...

Solutions

Calculus: An Applied Approach (Min...

Finding Present Value In Exercises 31-34, complete the table to determine the amount of money P that should be invested at rate r to produce a final balance of $100,000 in t years.

See Example 5.

n 1 2 4 12 365 Continuous compounding

A

r = 4.5 % , t = 30 years

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