Find the interest earned on an interest-bearing account

Question

Every month, a particular interest-bearing account earns 0.2 percent interest on the average balance for that month. The function B(t) =7.5t^2 -300t +5000 represents one investor's balance in this account during the month of November. B(t) gives the number of dollars in the account on day t, with t=0 being the beginning of the first day. How much interest will the account earn this month?

 

Answer 1

The average value of a function f(t) on an interval [a,b] is given by

∫_a^b f(t) dt /(b-a).

In your case, the average of B(t)=7.5t²-300t+500 on [0,30] is given by

∫₀³⁰ B(t) dt /30 = 2750.

0.2% interest on this balance is 2750*0.002 = $5.5.

Answer 2

The interest will be 0.003*<B(t)>, where <> means average.

 

<B(t)>=1/31*∫₀³¹B(t)dt=1/31*[55/30*31³-150*31²+5000*31]≈2111.83

 

Interest compounded will be 0.003*2111.83≈6.34

 

Answer: $6.34.

Answer 3

Hi Bella,

Can I assume that you have calculus for this? In order to calculate an average, I would integrate B(t) with respect to t, then divide it by the total time (31 days for July  30 days for November).

Int(5.5t²-300t+5000)dt = (5.5/3)t³-150t²+5000t |³¹₀ = $65,466.83

Int(7.5t²-300t+5000)dt = (7.5/3)t³-150t²+5000t |³⁰0 = $82,500

 

Average daily balance = $82,500/30 = $2,750.

The interest is 0.2% on the average balance, or

(0.002)($2,750) = $5.50

Answer 4

Dear Bella,

 

July has 31 days, so, if I'm reading the problem correctly

 

B(t) = (5.5)t² - 300t + 5000

 

Bella, I'm afraid there might be something wrong with this equation.  As written, it predicts that our poor account holder will be losing money.  For 0 > t ≤ 54.5455, 300*t will be greater than (5.5)*t² and the $5000 in the account will be diminished.  Not until day 55 arrives will the (5.5)t² term be larger than the (300)*t term.

 

Something doesn't flush her, Bella.  Have a look at that original equation again and see if you entered it correctly.