For the next 20 years, you plan to invest $600 a month in a stock account earning 7 percent annually and $400 a month in a bond account earning 4 percent per year. When you retire in 20 years, you will combine your money into an account with a return of 10 percent per year. How much can you withdraw each month during retirement assuming a 30-year withdrawal period?

Monthly withdraw= $204.55

Step-by-step explanation:

First, we need to calculate the value of the account at the moment of retirement using the following formula:

FV= {A*[(1+i)^n-1]}/i

A= monthly deposit

Stock:

Monthly deposit= $600

Interest rate= 0.07/12= 0.00583

Number of periods (n)= 20*12= 240 months

FV= {600*[(1.00583^240) - 1]} / 0.00583

FV= $312,404.24

Bond:

Monthly deposit= $400

Interest rate= 0.04/12= 0.0033

Number of periods (n)= 20*12= 240 months

FV= {400*[(1.0033^240) - 1]} / 0.0033

FV= $146,052.20

Total value at retirement= $458,456.44

Now, we can determine the monthly withdrawals after retirement:

Monthly withdraw= (FV*i) / [1 - (1+i)^(-n)]

FV= 458,456.44

n= 30*12= 360

i= 0.1/12= 0.0083

Monthly withdraw= (458,456.44*0.0083) / [(1.0083^360) - 1]

Monthly withdraw= $204.55

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

24+x=y

step-by-step explanation: