Question 7.2.7: Investment Strategies for Making a Million Money is invested...
Investment Strategies for Making a Million
Money is invested at 8% interest compounded continuously. If deposits are made continuously at the rate of $2000 per year, find the size of the initial investment needed to reach $1 million in 20 years.
Here, interest is earned at the rate of 8% and additional deposits are
assumed to be made on a continuous basis. If the deposit rate is $d per year, the amount A(t) in the account after t years satisfies the differential equation
\frac{d A}{d t}=0.08 A+d.
This equation is separable and can be solved by dividing both sides by 0.08 A + d and integrating. We have
\int \frac{1}{0.08 A+d} d A=\int 1 d t,so that
\frac{1}{0.08} \ln |0.08 A+d|=t+c .Using d = 2000, we have
12.5 \ln |0.08 A+2000|=t+c.
Setting t = 0 and taking A(0) = x gives us the constant of integration:
12.5 \ln |0.08 x+2000|=c,
so that
12.5 \ln |0.08 A+2000|=t+12.5 \ln |0.08 x+2000| \text {. } (2.7)
To find the value of x such that A(20) = 1,000,000, we substitute t = 20 and A = 1,000,000 into equation (2.7) to obtain
12.5 \ln |0.08(1,000,000)+2000|=20+12.5 \ln |0.08 x+2000|or 12.5 \ln |82,000|=20+12.5 \ln |0.08 x+2000|.
We can solve this for x, by subtracting 20 from both sides and then dividing by 12.5, to obtain
\frac{12.5 \ln 82,000-20}{12.5}=\ln |0.08 x+2000|.
Taking the exponential of both sides, we now have
e^{(12.5 \ln 82,000-20) / 12.5}=0.08 x+2000.
Solving this for x yields
x=\frac{e^{\ln 82,000-1.6}-2000}{0.08} \approx 181,943.93 .So, the initial investment needs to be $181,943.93 (slightly less than $200,000) in order to be worth $1 million at the end of 20 years.