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To solve this problem, we are given the equation relating the monthly income \( I \) (in thousands of soles) and the sales \( q \) (in hundreds of units) as follows:

\[
I^2 + 4I - 100q + 100 + 5q^3 = 0
\]

We need to find the differential of the income \( dI \) in terms of \( dq \) given that the current sales level is \( q = 200 \) units (i.e., \( q = 2 \) in hundreds).

### Steps:

1. **Differentiate the given equation implicitly** with respect to \( q \) to find \( \frac{dI}{dq} \) (denoted as \( dI/dq \)):
   \[
   \frac{d}{dq}(I^2 + 4I - 100q + 100 + 5q^3) = 0
   \]
   Differentiate each term:
   - \( \frac{d}{dq}(I^2) = 2I \frac{dI}{dq} \)
   - \( \frac{d}{dq}(4I) = 4 \frac{dI}{dq} \)
   - \( \frac{d}{dq}(-100q) = -100 \)
   - \( \frac{d}{dq}(100) = 0 \)
   - \( \frac{d}{dq}(5q^3) = 15q^2 \)

   Thus, the differentiated equation becomes:
   \[
   2I \frac{dI}{dq} + 4 \frac{dI}{dq} - 100 + 15q^2 = 0
   \]

2. **Solve for \( \frac{dI}{dq} \):**
   Combine terms:
   \[
   (2I + 4) \frac{dI}{dq} = 100 - 15q^2
   \]

   Now solve for \( \frac{dI}{dq} \):
   \[
   \frac{dI}{dq} = \frac{100 - 15q^2}{2I + 4}
   \]

3. **Substitute \( q = 2 \):**
   The problem states that the current sales level is 200 units, which means \( q = 2 \) (since sales are in hundreds).

   So, substituting \( q = 2 \) into the equation:
   \[
   \frac{dI}{dq} = \frac{100 - 15(2)^2}{2I + 4}
   \]
   Simplifying:
   \[
   \frac{dI}{dq} = \frac{100 - 15(4)}{2I + 4} = \frac{100 - 60}{2I + 4} = \frac{40}{2I + 4}
   \]

4. **Find \( I \) at \( q = 2 \):**
   To find \( I \) when \( q = 2 \), substitute \( q = 2 \) into the original equation:
   \[
   I^2 + 4I - 100(2) + 100 + 5(2)^3 = 0
   \]
   Simplifying:
   \[
   I^2 + 4I - 200 + 100 + 5(8) = 0
   \]
   \[
   I^2 + 4I - 100 + 40 = 0
   \]
   \[
   I^2 + 4I - 60 = 0
   \]

   Solve this quadratic equation using the quadratic formula:
   \[
   I = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(-60)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm \sqrt{256}}{2} = \frac{-4 \pm 16}{2}
   \]
   Therefore, the two possible solutions for \( I \) are:
   \[
   I = \frac{-4 + 16}{2} = 6 \quad \text{or} \quad I = \frac{-4 - 16}{2} = -10
   \]

5. **Choose the appropriate \( I \):**
   Since \( I \) represents income, it cannot be negative. Thus, we take \( I = 6 \).

6. **Substitute \( I = 6 \) back into the differential equation:**
   Now that we know \( I = 6 \), substitute it into the expression for \( \frac{dI}{dq} \):
   \[
   \frac{dI}{dq} = \frac{40}{2(6) + 4} = \frac{40}{12 + 4} = \frac{40}{16} = 2.5
   \]

Thus, the differential of the income is:

\[
dI = 2.5 \, dq
\]

The correct answer is **\( dI = 2.5 \, dq \)**.

---

Would you like more details or have any other questions?

Here are 5 related questions to explore:
1. How does the implicit differentiation process work in general?
2. What are some common applications of differential equations in economics?
3. What would the outcome be if the sales level \( q \) increased to 3?
4. How can the quadratic formula be used to solve real-world business problems?
5. What is the significance of finding the correct solution for \( I \) in this context?

**Tip**: When solving implicit differentiation problems, always remember to apply the chain rule correctly when differentiating terms with more than one variable!

Los ingresos mensuales I, de una empresa (en miles de soles), después de vender q cientos de sus artículos, vienen relacionados por: I^2 + 4I - 100q + 100 + 5q^3 = 0. Si actualmente el nivel de ventas es de 200 unidades, entonces el diferencial del ingreso (dI) resulta:

Learn how to find the differential of income (dI) using implicit differentiation from the equation I^2 + 4I - 100q + 100 + 5q^3 = 0, given that the sales level is 200 units. This problem involves calculus concepts such as implicit differentiation and solving quadratic equations.

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