A company is producing and selling a product, with the production cost per unit being \$10 and the selling price per unit being \$15. The company's initial production rate is 100 units per month, and it is increasing by 10% every month. The company's total revenue can be modeled by the differential equation $\frac{dR}{dt} = (15 - 10) \cdot (100 \cdot 1.1^t)$, where R is the total revenue and t is the time in months. Find the total revenue after 6 months, rounded to the nearest dollar.

The differential equation for the company's revenue is given by:

\[
\frac{dR}{dt} = (15 - 10) \cdot (100 \cdot 1.1^t)
\]

Since the selling price per unit is $15 and the cost per unit is $10, the profit per unit is $15 - 10 = 5$. Thus, we have:

\[
\frac{dR}{dt} = 5 \cdot (100 \cdot 1.1^t) = 500 \cdot 1.1^t
\]

To find the total revenue $ R(t) $ after 6 months, we need to integrate this differential equation:

\[
R(t) = \int 500 \cdot 1.1^t \, dt
\]

### Step 1: Integrate $ 500 \cdot 1.1^t $

The integral of $ 1.1^t $ is calculated as follows:
\[
\int 1.1^t \, dt = \frac{1.1^t}{\ln(1.1)}
\]

Therefore,
\[
R(t) = 500 \cdot \frac{1.1^t}{\ln(1.1)} + C
\]

### Step 2: Solve for the constant $ C $

Since $ R(0) = 0 $ (assuming revenue is zero at $ t = 0 $), we substitute $ t = 0 $ and $ R = 0 $ to find $ C $:

\[
0 = 500 \cdot \frac{1.1^0}{\ln(1.1)} + C \Rightarrow C = -\frac{500}{\ln(1.1)}
\]

Thus, the total revenue function becomes:
\[
R(t) = 500 \cdot \frac{1.1^t}{\ln(1.1)} - \frac{500}{\ln(1.1)}
\]

### Step 3: Calculate $ R(6) $

Now, substitute $ t = 6 $:

\[
R(6) = 500 \cdot \frac{1.1^6}{\ln(1.1)} - \frac{500}{\ln(1.1)}
\]

### Numerical Calculation

1. Calculate $ 1.1^6 \approx 1.771561 $
2. Using $ \ln(1.1) \approx 0.09531 $:

\[
R(6) \approx 500 \cdot \frac{1.771561}{0.09531} - \frac{500}{0.09531}
\]
\[
\approx 500 \cdot 18.586 - 500 \cdot 10.493
\]
\[
\approx 9293 - 5246.5 = 4046.5
\]

Rounding to the nearest dollar, the total revenue after 6 months is approximately:

\[
\boxed{4047}
\]

Would you like further details on any step, or have any other questions?

---

1. What would happen to the revenue if the production rate increased by 20% instead of 10%?
2. How does the revenue change if the initial production rate was 200 units?
3. What if the profit per unit was different, say $7 per unit?
4. How would we model total costs and profits, not just revenue?
5. What would be the revenue after 12 months?

**Tip:** Differential equations can model dynamic changes in business metrics, such as revenue or costs over time.

A company is producing and selling a product, with the production cost per unit being $10 and the selling price per unit being $15. The company's initial production rate is 100 units per month, and it is increasing by 10% every month. The company's total revenue can be modeled by the differential equation $ \frac{dR}{dt} = (15 - 10) \cdot (100 \cdot 1.1^t) $, where R is the total revenue and t is the time in months. Find the total revenue after 6 months, rounded to the nearest dollar.

Learn how to calculate the total revenue after 6 months for a company using differential equations and integration. This problem involves concepts of exponential growth and revenue modeling, suitable for high school and college students. Initial production rate is 100 units per month, increasing by 10%, with a unit profit of $5.

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