Solution A small town decides to conduct a fund-raising drive for a new park facilities for kids. The cost is $50 000. The initial amount given by City Hall is $8 000. The contribution to the fund F is increasing at a rate proportional to the difference between the cost of $50 000 and the amount F at time t in months. After one month, $36 000 is in the fund. How much will be in the fund after 3 months

The fund, F, after 3 months = $48,442.25

Step-by-step explanation:

Amount to be made = $50000

Initial amount = F₀ = $8000

The contribution to the fund F is increasing at a rate proportional to the difference between the cost of $50,000 and the amount F at time t in months.

Noting that F ≤ 50,000

(dF/dt) ∝ (50,000 - F)

k = constant of proportionality

F' = k(50000 - F)

dF/(50000 - F) = k dt

∫ dF/(50000 - F) = ∫ kdt

- In [50,000 - F] = kt + c

c = constant of integration

At t = 0, F = $8000

- In [50,000 - 8,000] = 0 + c

c = - In 42,000 = - 10.645

- In [50,000 - F] = kt - 10.645

After one month, $36,000 is in the fund.

at t = 1 month, F = $36,000

- In [50,000 - 36,000] = k - 10.645

k - 10.645 = - In 14,000 = - 9.547

k = 10.645 - 9.547 = 1.098

- In [50,000 - F] = 1.098t - 10.645

In [50,000 - F] = 10.645 - 1.098t

At t = 3 months,

In [50,000 - F] = 10.645 - 1.098(3) = 7.351

50,000 - F = e⁷•³⁵¹ = 1557.7535

F = 50,000 - 1557.7535 = $48,442.2465

The fund, F, after 3 months = $48,442.25

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step-by-step explanation: