Let's break the problem into two parts: --- ### **Part 1: Accumulated Amount After 3 Years (Cost = \$200,000)** #### **Given:** - Income function: \( f(t) = 75000e^{0.06t} \) - Market rate of interest: 7% (0.07) - Time \( t \) in years, \( t = 0 \) to \( t = 3 \) - Need to check if accumulated value at \( t = 3 \) covers \$200,000 #### **Formula for Accumulated Value with Continuous Compounding:** \[ A = \int_{0}^{3} f(t) \cdot e^{r(3-t)} \, dt \] where \( r = 0.07 \) (market rate). #### **Plug values into the formula:** \[ A = \int_{0}^{3} 75000e^{0.06t} \cdot e^{0.07(3-t)} \, dt \] \[ = \int_{0}^{3} 75000e^{0.06t} \cdot e^{0.21 - 0.07t} \, dt \] \[ = \int_{0}^{3} 75000e^{0.21} e^{0.06t - 0.07t} \, dt \] \[ = 75000e^{0.21} \int_{0}^{3} e^{-0.01t} \, dt \] #### **Integrate:** \[ \int e^{-0.01t} dt = -100e^{-0.01t} \] So, \[ A = 75000e^{0.21} \left[ -100e^{-0.01t} \right]_{0}^{3} \] \[ = 75000e^{0.21} \left(-100e^{-0.03} + 100e^{0}\right) \] \[ = 75000e^{0.21} \left(100(1 - e^{-0.03})\right) \] #### **Calculate values:** - \( e^{0.21} \approx 1.2337 \) - \( e^{-0.03} \approx 0.9704 \) So, \[ A = 75000 \times 1.2337 \times 100 \times (1 - 0.9704) \] \[ = 75000 \times 1.2337 \times 100 \times 0.0296 \] \[ = 75000 \times 1.2337 \times 2.96 \] \[ = 75000 \times 3.6522 \approx \$273,915 \] **Conclusion:** The accumulated value (\$273,915) is greater than \$200,000. The company can cover the machine cost. --- ### **Part 2: Accumulated Amount for Years 4-8 (Machine Cost = \$30,000)** #### **Given:** - New annual income: \$80,000 growing at 4% per year - \( f(t) = 80000e^{0.04t} \) (for \( t = 0 \) to \( t = 4 \); this is years 4 to 8 from today!) - Market interest rate: 5% (0.05) #### **Accumulated Value at end of year 8:** \[ A = \int_{0}^{4} 80000e^{0.04t} \cdot e^{0.05(4-t)} dt \] \[ = \int_{0}^{4} 80000e^{0.04t} \cdot e^{0.20 - 0.05t} dt \] \[ = 80000e^{0.20} \int_{0}^{4} e^{0.04t - 0.05t} dt \] \[ = 80000e^{0.20} \int_{0}^{4} e^{-0.01t} dt \] Integrate: \[ \int_{0}^{4} e^{-0.01t} dt = -100e^{-0.01t} \Big|_{0}^{4} = -100(e^{-0.04} - 1) \] Calculate: - \( e^{0.20} \approx 1.2214 \) - \( e^{-0.04} \approx 0.9608 \) So, \[ A = 80000 \times 1.2214 \times 100 \times (1 - 0.9608) \] \[ = 80000 \times 1.2214 \times 3.92 \] \[ = 80000 \times 4.7879 \approx \$383,032 \] **Conclusion:** The accumulated amount (\$383,032) is much greater than \$30,000. The company can purchase the new machine. --- ## **Final Answers:** 1. **After 3 years, the accumulated value is approximately \$273,915, which covers the \$200,000 machine cost.** 2. **After years 4-8, the accumulated value is approximately \$383,032, which covers the \$30,000 machine cost.** If you need further breakdown or clarification, let me know!

Certainly! Let's carefully go through each part step-by-step, detailing all calculations and reasoning. --- ## **Part 1: Will the company's earnings over 3 years cover the \$200,000 cost of the machine?** ### **Given Data:** - Income function: \(\displaystyle f(t) = 75,000 e^{0.06t}\) - Discount rate (market interest): \(\displaystyle r = 7\% = 0.07\) - Time period: \(t = 0\) to \(t=3\) years - Cost of machine: \$200,000 --- ### **Step 1: Understand what is asked** We need to compute the **present value** of the income flow over 3 years, considering continuous compounding at the market rate, and determine if this amount covers \$200,000. Since income flows occur over time, we use the **present value of a continuous income stream**, which is: \[ PV = \int_{0}^{T} f(t) \cdot e^{r(T - t)} dt \] Here, \(T=3\). --- ### **Step 2: Write the integral** \[ PV = \int_{0}^{3} f(t) \cdot e^{r(3 - t)} dt \] Substitute \(f(t) = 75,000 e^{0.06t}\): \[ PV = \int_{0}^{3} 75,000 e^{0.06t} \cdot e^{0.07(3 - t)} dt \] --- ### **Step 3: Simplify the exponential terms** \[ PV = 75,000 \int_{0}^{3} e^{0.06t} \cdot e^{0.21 - 0.07t} dt \] \[ = 75,000 \int_{0}^{3} e^{0.21} \cdot e^{0.06t - 0.07t} dt \] \[ = 75,000 e^{0.21} \int_{0}^{3} e^{-0.01t} dt \] --- ### **Step 4: Factor out constants and compute the integral** Recall: \[ \int e^{ax} dx = \frac{1}{a} e^{ax} \] So, \[ \int_{0}^{3} e^{-0.01t} dt = \left[ -\frac{1}{0.01} e^{-0.01t} \right]_0^3 = -100 \left(e^{-0.03} - 1 \right) \] --- ### **Step 5: Calculate the numerical values** - \( e^{0.21} \approx 1.2337 \) - \( e^{-0.03} \approx 0.9704 \) Now, \[ PV = 75,000 \times 1.2337 \times (-100) \times (0.9704 - 1) \] Note that: \[ 0.9704 - 1 = -0.0296 \] Thus, \[ PV = 75,000 \times 1.2337 \times (-100) \times (-0.0296) \] The negatives cancel: \[ PV = 75,000 \times 1.2337 \times 100 \times 0.0296 \] --- ### **Step 6: Final calculation** Calculate step by step: - \(75,000 \times 1.2337 \approx 92,527.5\) - Multiply by 100: \(92,527.5 \times 100 = 9,252,750\) - Multiply by 0.0296: \[ 9,252,750 \times 0.0296 \approx 273,915 \] **Therefore,** \[ \boxed{ \text{Present value of income over 3 years} \approx \boxed{\$273,915} } \] --- ### **Conclusion for Part 1:** Since \$273,915 > \$200,000, **the company's earnings over 3 years are sufficient to cover the machine cost**. --- ## **Part 2: Will earnings from years 4 to 8 cover the \$30,000 cost of a new machine?** ### **Given Data:** - New annual income: \(f(t) = 80,000 e^{0.04t}\) - Time period: years 4 to 8, i.e., \(t = 4 \text{ to } 8\) (from today’s perspective, but for calculations, we treat \(t=0\) as start of this period) - Discount rate: \(r = 5\% = 0.05\) ### **Step 1: Reframe the problem** We want the **accumulated value at the end of year 8** of the income during years 4–8, considering the discount rate. Since the income starts at year 4, and we want the **value at year 8**, we can model the income flow over \(t=0\) to \(t=4\) (corresponding to years 4–8), and then discount it back to year 8 or directly compute the accumulation at year 8. **Approach:** Compute the **present value at year 4** of the income during years 4–8, then accumulate that to year 8. But a more straightforward approach is: \[ A = \int_{0}^{4} f(t) \cdot e^{r(4 - t)} dt \] This gives the **value at year 4** of income from years 4 to 8. Then, to get the amount **at year 8**, we multiply by \(e^{r(4)}\), since from year 4 to year 8 is 4 years, and we discount or compound accordingly. Alternatively, since the income occurs during years 4–8, and the question asks for the accumulated amount **during** that period, it's best to find the total accumulated at year 8. --- ### **Step 2: Write the integral for income during years 4–8** Using the **discounted cash flow**: \[ A_{8} = \int_{t=0}^{4} f(t) \cdot e^{r(8 - t)} dt \] or, equivalently, the accumulated value at year 8 of income received during years 4–8. Substituting \(f(t) = 80,000 e^{0.04t}\): \[ A_{8} = \int_{0}^{4} 80,000 e^{0.04t} \cdot e^{0.05(8 - t)} dt \] Simplify: \[ A_{8} = 80,000 \int_{0}^{4} e^{0.04t} \cdot e^{0.4 - 0.05t} dt \] \[ = 80,000 e^{0.4} \int_{0}^{4} e^{0.04t - 0.05t} dt \] \[ = 80,000 e^{0.4} \int_{0}^{4} e^{-0.01 t} dt \] --- ### **Step 3: Compute the integral** \[ \int_{0}^{4} e^{-0.01 t} dt = -100 e^{-0.01 t} \Big|_{0}^{4} = -100 (e^{-0.04} - 1) \] Calculate: - \( e^{0.4} \approx 1.4918 \) - \( e^{-0.04} \approx 0.9608 \) Now, \[ A_{8} = 80,000 \times 1.4918 \times (-100) \times (0.9608 - 1) \] Note: \[ 0.9608 - 1 = -0.0392 \] So, \[ A_{8} = 80,000 \times 1.4918 \times (-100) \times (-0.0392) \] Negatives cancel: \[ A_{8} = 80,000 \times 1.4918 \times 100 \times 0.0392 \] --- ### **Step 4: Calculate step-by-step** - \(80,000 \times 1.4918 \approx 119,344\) - Multiply by 100: \(119,344 \times 100 = 11,934,400\) - Multiply by 0.0392: \[ 11,934,400 \times 0.0392 \approx 468,304 \] **Therefore,** \[ \boxed{ \text{Accumulated amount at year 8} \approx \$468,304 } \] which is **much greater** than the \$30,000 needed to purchase the new machine outright. --- ## **Final Summary:** | Part | Estimated Amount | Conclusion | |---------|-----------------------|--------------| | **Part 1** | **\$273,915** | Income over 3 years exceeds \$200,000; sufficient to cover the existing machine. | | **Part 2** | **\$468,304** | Income during years 4–8 exceeds \$30,000; sufficient to buy the new machine outright. | --- If you need further clarification or detailed steps, feel free to ask!

The accumulated income over 3 years is approximately \$273,915, which exceeds the \$200,000 cost of the machine, so the company can cover the existing machine's cost. The accumulated income during years 4 to 8 is approximately \$468,304, which is sufficient to purchase the new machine outright at \$30,000.