Question

Stacey and her husband David have a joint savings account that earns 3.5% interest payable continuously and has a current balance of $58,458. Each year, David wishesto withdraw $4000 payable continuously at a level rate. Stacey wishes to deposit X at the beginning of each year (for thirty years) so that the account will last for30 years. What is the least X that will work?

Answer 1

The present value of withdrawals is an annuity with payment = 4000, interest rate = 3.5% and number of years = 30. On computing, the present value =$73,568.18

The savings account has a current balance of $58,458
Therefore the additional payments deposited need to generate a present value of only 73,568.18 - 58,458 = $15,110.18
On calculating, the yearly payments for the annuity = $821.56

Therefore $821.56 is the least X that will work.

Answer 2

PV of amount withdrawn = PV of amount deposited
4000/0.035=58458+(X/0.035)(1-1/1.035^30)
114285.71=58458+(X/0.035)*0.6437
55827.71=18.3914X
X=3035.53

Answer 3

Oh, I am sorry, I missed that the interest rate is continuously compounded.

So with yearly withdrawals of $4,000, the present value would be
4000/(e^0.035) + 4000/(e^(0.035*2)) + ... + 4000/(e^(0.035*30)) = $73,000.29

The current account balance = $58,458
Hence, the present value of deposits needs to be equal to = 73000.29-58458 = $14,542.29
The payments are made at the beginning of each year, hence the present value would be equal to
X + X/(e^0.035) + ...+ X / (e^(0.035*29))
Solving for X, the yearly payments can be computed as $769.43

Answer 4

4000*((1-e^(-.o35*30))/.035)=74292.82867 .035=ln(1+i) => i=.0356197088 X*((1-(1.0356197088)^(-30))/.0343945837)=74292.82867-58458 Basically: X*(annuitydue)=15834.82867 and then X=837.8156707=837.82

Answer 5

huh idk it