Let’s solve the problem step by step: --- ### **Given:** - Cost in 1962 = $0.05 - Cost in 2010 = $0.45 - Exponential growth: \( S(t) = S_0 e^{kt} \), where \( S_0 \) is the initial value, and \( k \) is the growth rate. - Let \( t = 0 \) at 1964. --- ### **a) Write the function \( S \) for the cost of a stamp \( t \) years after 1964** #### **Step 1: Find \( S_0 \) for 1964** Since we have 1962 and 2010, but \( t=0 \) is 1964, we need to get the value for 1964. Let’s assume the cost in 1964 is \( x \). From 1962 to 2010 is 48 years, cost increases from $0.05 to $0.45. Let’s use the exponential formula: \[ S(t) = S_0 e^{kt} \] Let’s set **1962 as \( t = -2 \)** and **2010 as \( t = 46 \)** (since 1964 is \( t = 0 \)). So: - \( S(-2) = 0.05 \) - \( S(46) = 0.45 \) Assume \( S(t) = S_0 e^{kt} \). We have: \[ \begin{align*} S(-2) &= S_0 e^{-2k} = 0.05 \\ S(46) &= S_0 e^{46k} = 0.45 \\ \end{align*} \] Divide these equations: \[ \frac{S(46)}{S(-2)} = \frac{S_0 e^{46k}}{S_0 e^{-2k}} = e^{48k} = \frac{0.45}{0.05} = 9 \] \[ e^{48k} = 9 \] \[ 48k = \ln(9) \] \[ k = \frac{\ln(9)}{48} \] Now, plug back to find \( S_0 \) (the cost in 1964): \[ S(-2) = S_0 e^{-2k} = 0.05 \implies S_0 = 0.05 e^{2k} \] --- ### **b) What was the growth rate in the cost?** \[ k = \frac{\ln(9)}{48} = \frac{2.1972}{48} \approx 0.04577 \] So, **growth rate** = \( \boxed{0.0458} \) per year (rounded to 4 decimal places). --- ### **a, continued: Write function \( S(t) \) fully** \[ S_0 = 0.05 e^{2k} = 0.05 e^{2 \times 0.04577} = 0.05 e^{0.09154} \approx 0.05 \times 1.0958 \approx 0.0548 \] So, \[ \boxed{S(t) = 0.0548 \, e^{0.0458 t}} \] where \( t \) is years after 1964. --- ### **c) Predict the cost in 2017, 2020, and 2023** - \( t_{2017} = 2017 - 1964 = 53 \) - \( t_{2020} = 2020 - 1964 = 56 \) - \( t_{2023} = 2023 - 1964 = 59 \) Calculate each: #### **For 2017:** \[ S(53) = 0.0548 \times e^{0.0458 \times 53} = 0.0548 \times e^{2.4274} \] \[ e^{2.4274} \approx 11.33 \] \[ S(53) \approx 0.0548 \times 11.33 \approx 0.620 \text{ dollars } = 62.0 \text{ cents} \] #### **For 2020:** \[ S(56) = 0.0548 \times e^{0.0458 \times 56} = 0.0548 \times e^{2.5648} \] \[ e^{2.5648} \approx 12.99 \] \[ S(56) \approx 0.0548 \times 12.99 \approx 0.713 \text{ dollars } = 71.3 \text{ cents} \] #### **For 2023:** \[ S(59) = 0.0548 \times e^{0.0458 \times 59} = 0.0548 \times e^{2.7022} \] \[ e^{2.7022} \approx 14.92 \] \[ S(59) \approx 0.0548 \times 14.92 \approx 0.817 \text{ dollars } = 81.7 \text{ cents} \] --- ### **d) How much will the firm save by buying Forever stamps in 2010?** - In 2010, firm buys 10,000 Forever Stamps for $4,500 (i.e., $0.45 each). - Needs 10,000 first-class stamps each year from 2010 to 2024 (15 years). - If buying yearly, would pay the price predicted for each year. **We need to find the total cost if bought yearly, using the predicted prices (including the jumps in 2017, 2020, and 2023):** Let’s calculate the cost for each year, round to nearest cent per stamp, multiply by 10,000, and sum for 2010–2024. #### **Stamps' Predicted Costs (rounded to nearest cent):** - 2010: \( t = 46 \) \( S(46) = 0.0548 \times e^{0.0458 \times 46} = 0.0548 \times e^{2.1068} \approx 0.0548 \times 8.217 \approx 0.451 \) → **$0.45** - 2011: \( t = 47 \) \( S(47) = 0.0548 \times e^{0.0458 \times 47} \approx 0.471 \) → **$0.47** - 2012: \( S(48) \approx 0.493 \) → **$0.49** - 2013: \( S(49) \approx 0.515 \) → **$0.52** - 2014: \( S(50) \approx 0.539 \) → **$0.54** - 2015: \( S(51) \approx 0.564 \) → **$0.56** - 2016: \( S(52) \approx 0.590 \) → **$0.59** - 2017: \( S(53) \approx 0.620 \) → **$0.62** - 2018: \( S(54) \approx 0.646 \) → **$0.65** - 2019: \( S(55) \approx 0.679 \) → **$0.68** - 2020: \( S(56) \approx 0.713 \) → **$0.71** - 2021: \( S(57) \approx 0.747 \) → **$0.75** - 2022: \( S(58) \approx 0.781 \) → **$0.78** - 2023: \( S(59) \approx 0.817 \) → **$0.82** - 2024: \( S(60) \approx 0.855 \) → **$0.86** #### **Total cost if bought yearly:** \[ \begin{align*} 2010: & \ 10,000 \times 0.45 = 4,500 \\ 2011: & \ 10,000 \times 0.47 = 4,700 \\ 2012: & \ 10,000 \times 0.49 = 4,900 \\ 2013: & \ 10,000 \times 0.52 = 5,200 \\ 2014: & \ 10,000 \times 0.54 = 5,400 \\ 2015: & \ 10,000 \times 0.56 = 5,600 \\ 2016: & \ 10,000 \times 0.59 = 5,900 \\ 2017: & \ 10,000 \times 0.62 = 6,200 \\ 2018: & \ 10,000 \times 0.65 = 6,500 \\ 2019: & \ 10,000 \times 0.68 = 6,800 \\ 2020: & \ 10,000 \times 0.71 = 7,100 \\ 2021: & \ 10,000 \times 0.75 = 7,500 \\ 2022: & \ 10,000 \times 0.78 = 7,800 \\ 2023: & \ 10,000 \times 0.82 = 8,200 \\ 2024: & \ 10,000 \times 0.86 = 8,600 \\ \end{align*} \] Sum up: \[ \begin{align*} 4,500 + 4,700 + 4,900 + 5,200 + 5,400 + 5,600 + 5,900 + 6,200 + 6,500 + 6,800 \\ + 7,100 + 7,500 + 7,800 + 8,200 + 8,600 = 99,900 \end{align*} \] #### **Total cost if buy Forever Stamps in 2010:** \[ 15 \text{ years} \times 10,000 \times \$0.45 = 150,000 \times 0.45 = \$67,500 \] But according to the question, the firm spends **$4,500 for 10,000 stamps** in 2010, and uses them for each year (so for each year, it would have paid $4,500 instead of the higher price). #### **How much does the firm save?** \[ \text{Total savings} = \$99,900 - (\$4,500 \times 15) = \$99,900 - \$67,500 = \$32,400 \] --- ### **Final Answers:** **a)** \[ \boxed{S(t) = 0.0548 \, e^{0.0458 t}} \] **b)** \[ \boxed{\text{Growth rate} = 0.0458 \text{ per year } (4.58\%)} \] **c)** \[ \begin{align*} 2017: & \quad \boxed{\$0.62} \\ 2020: & \quad \boxed{\$0.71} \\ 2023: & \quad \boxed{\$0.82} \end{align*} \] **d)** \[ \boxed{\$32,400} \] **(The firm saves $32,400 by buying Forever Stamps at the start of 2010 rather than paying the predicted yearly prices from 2010 to 2024.)**