Question: Suppose that $63,000 is invested at 6 1/2% interest, compounded quarterly. a) Find the function…

Suppose that $63,000 is invested at 6 1/2% interest, compounded quarterly.

a) Find the function for the amount to which the investment grows after t years.

b) Graph the function.

c) Find the amount of money in the account at t = 0, 2, 8, and 10 years.

The amount of money in the account at t = 0 years is $63,000.

The amount of money in the account at t = 2 years is $71,671.

The amount of money in the account at t = 8 years is $105,526.

The amount of money in the account at t = 10 years is $120,050.

d) When will the amount of money in the account reach $300,000?

Solution

The problem involves compound interest. Let’s solve each part step by step. a) Find the function for the amount to which the investment grows after \( t \) years. The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after \( n \) years, including interest. - \( P \) is the principal amount (initial investment), \( P = \$63,000 \). - \( r \) is the annual interest rate (decimal), \( r = 0.065 \). - \( n \) is the number of times that interest is compounded per year, \( n = 4 \) (quarterly). - \( t \) is the number of years. The function becomes: \[ A(t) = 63000 \left(1 + \frac{0.065}{4}\right)^{4t} \] c) Find the amount of money in the account at \( t = 0, 2, 8, \) and \( 10 \) years. 1. For \( t = 0 \): \[ A(0) = 63000 \left(1 + \frac{0.065}{4}\right)^{4 \times 0} = 63000 \] 2. For \( t = 2 \): \[ A(2) = 63000 \left(1 + \frac{0.065}{4}\right)^{4 \times 2} \approx 71671 \] 3. For \( t = 8 \): \[ A(8) = 63000 \left(1 + \frac{0.065}{4}\right)^{4 \times 8} \approx 105526 \] 4. For \( t = 10 \): \[ A(10) = 63000 \left(1 + \frac{0.065}{4}\right)^{4 \times 10} \approx 120050 \] d) When will the amount of money in the account reach $300,000? Solve for \( t \) when \( A(t) = 300000 \): \[ 300000 = 63000 \left(1 + \frac{0.065}{4}\right)^{4t} \] Divide both sides by 63000: \[ \frac{300000}{63000} = \left(1 + \frac{0.065}{4}\right)^{4t} \] \[ 4.7619 = \left(1.01625\right)^{4t} \] Take the natural logarithm of both sides: \[ \ln(4.7619) = 4t \cdot \ln(1.01625) \] Solve for \( t \): \[ t = \frac{\ln(4.7619)}{4 \cdot \ln(1.01625)} \approx 27.29 \] Thus, it will take approximately 27.29 years for the account to grow to $300,000.