Question

We are investigating two different retirment plans that we can invest our money in. Plan A invests $45,000 with an annual interest rate of 4.8% compounded continuously, and plan B invests $50,000 with an annual interest rate of 4.5%

Suppose we plan on retiring in 5 years. how much will plan A be worth at that time?

How much will Plan B be worth in 5 years?

Suppose instead we need at least $100,000 in this account we retire, and we don't plan on contributing any more to this account. (note that you actually need quite a bit more to retire, we'll assume that we have money in other accounts as well). How long will it take plan A to reach a balance of 100,000?

How long will it take plan B to reach a balance of $100,000?

Using your graphing calculator to graph the compound interest function for Plan A and the compund interest for Plan B using the window setting [0, 45] by [0, 400000]. Find the point of intersection (rounded to two decimal places) between the two functions, and interpert it.

when would it make more sense to choose plan A? (hint: When would plan A be worth more then plan B?)

When would it make more sense to choose Plan B?

Fine the T value of the point of interesction that you found in (what would it make more sense to choose Plan B) Algebraically. Show all work and incomplete solutions recieve no credit do not round answer til the end of this problem.

Please show all work for each one please Thank you so much!

Answer 1

1. The formula for continuous compounding is A = Pe^rt, where P is the principal/initial amount, A is the maturity value, r is the rate of interest and t is the no. of years. In plan A, we have A = 45000e^0.048*5 = 45000e^0.24 = 45000*1.27124915 = $ 57206.21 ( on rounding off to the nearest cent). The formula for compound interest is A = P(1+r)ⁿ. In plan B, the maturity value is 45000(1+ 4.8/100)⁵=45000*(1.048)⁵ = 45000*1.264172717 = $56887.77.

2. For plan A, let it take t more years to reach $ 100000. Then 100000 = 57206.21e^0.048t or, e^0.048t = 100000/57206.21 = 1.748061967. Now, on taking natural log of both the sides, we have 0.048t = ln1.748061967 = 0.558507726. Hence, t = 0.558507726/0.048 = 11.64 years. In case of plan B, 100000 =56887.77(1+4.8/100)^t or, (1.048)^t = 100000/56887.77=1.757847073. Now, on taking log of both the sides, we have t log 1.048 = log1.757847073 so that t = log1.757847073/log 1.048 = 0.24498109/ 0.020361282 = 12.03 years.

Graphs of plan A (in red) and plan B( in blue) are attached. The 2 graphs intersect at only one point where time (in years) is 0 i.e. at the starting time. It implies that the amount in case of continuous compounding will always be higher than that in case of compounding of interest. If the rate of interest is x ( in decimals) and the time is t years, e^xt will always be more than (1+x)^t . Thus, it will always make more sense to choose plan A.