Fund F accumulates at a rate of forceinterestt 11t for t 0

Fund F accumulates at a rate of: force_interest_t = 1/(1+t) for t > 0.
Fund G accumulates a rate of: force_interest_t = 4t/(1+2t^2) for t > 0.

You are given the following pieces of information:

(i) F(t) = Amount in Fund F at time t

(ii) G(t) = Amount in Fund G at time t

(iii) H(t) = F(t) – G(t)

(iv) F(0) = G(0)

(v) T is the time t when H(t) is a maximum.

Calculate T.

i) F(t) = integral 1/(1+t) dt = ln(1+t) + C

ii) G(t) = integral 4t/(1+2t^2) dt

Let ,(1+2t^2) =u

=> 4tdt = du --------------------Replace in original equation

G(t) = integral 1/u du

= ln(u)

= ln (1+2t^2)) +D

iii) H(t) = F(t) – G(t)

= ln(1+t) + C - ln (1+2t^2) - D

= ln(1+t) - ln (1+2t^2) + C - D

iv)F(0) = G(0)

=> ln(1+0) + C = ln (1+2*0^2) + D

=> C = D

v)For Maximum H(t) ,

H(t) =ln(1+t) - ln (1+2t^2) + C - D

H \'(t) =0

H\'(t) = 1/(1+t) - 4t/(1+2t^2) =0

=> 1/(1+t) = 4t/(1+2t^2)

=> 1+2t^2 = 4t + 4t^2

=> 2t^2+4t-1 =0

=> t = (-4+sqrt(16-4*2*(-1)) / (2*2) = 0.2247

Therefore, T = 0.2247 Answer