Answer:
Step-by-step explanation:
Given that a young person with no initial capital invests k dollars per year at an annual rate of return r. Assume that investments are made continuously and that the return is compounded continuously
a) Amount accumulated at time t
[tex]S(t) = ke^{rt}[/tex], where r = rate of interest and t = time lapsed
b) Here S = 1,000,000
[tex]1000000 = ke^{0.075(40)} \\k= 1000000e^{-0.075(40)}\\=49802.56[/tex]
So 49802.56 dollars to be invested to get 1 million after 40 years
c) k = 2000 per year
I 2000 will be for 40 years, II 2000 for 39 years, ..... Last 2000 for 0 years
i.e. final amount would be
[tex]2000[(e^{40r} +(e^{39r} +....e^r] = 1000000\\500 = \frac{e^r(e^{40r}-1}{e^r-1}[/tex]
solving we get r = 9.6% approxy