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?

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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.