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### Problem 1: Depreciation of Agricultural Machinery
The annual depreciation of agricultural machinery follows a linear model. Given:
- Annual depreciation = ₡1,750,000
- Initial value = ₡52,800,000

We need to answer the following:

#### a) Equation of the depreciation function

Let \( V(t) \) be the value of the machinery after \( t \) years, and since the depreciation is linear, the equation can be written as:
\[
V(t) = V_0 - D \cdot t
\]
Where:
- \( V_0 = 52,800,000 \) is the initial value.
- \( D = 1,750,000 \) is the annual depreciation.

Thus, the depreciation function is:
\[
V(t) = 52,800,000 - 1,750,000 \cdot t
\]

#### b) Useful life of the machinery

The useful life \( L \) of the machinery is the time it takes for the value to reach zero. We set \( V(t) = 0 \) and solve for \( t \):

\[
52,800,000 - 1,750,000 \cdot t = 0
\]
\[
1,750,000 \cdot t = 52,800,000
\]
\[
t = \frac{52,800,000}{1,750,000} = 30.17 \text{ years}
\]

Therefore, the useful life of the machinery is approximately **30 years**.

#### c) After how many years will the value be ₡18,000,000?

We need to solve for \( t \) when \( V(t) = 18,000,000 \):

\[
52,800,000 - 1,750,000 \cdot t = 18,000,000
\]
\[
1,750,000 \cdot t = 52,800,000 - 18,000,000
\]
\[
1,750,000 \cdot t = 34,800,000
\]
\[
t = \frac{34,800,000}{1,750,000} = 19.89 \text{ years}
\]

So, the value will be ₡18,000,000 after approximately **19.9 years**.

#### d) How much has this equipment depreciated after 4 years?

To find the depreciation after 4 years, substitute \( t = 4 \) into the depreciation function:
\[
V(4) = 52,800,000 - 1,750,000 \cdot 4
\]
\[
V(4) = 52,800,000 - 7,000,000 = 45,800,000
\]

Thus, the equipment's value after 4 years is ₡45,800,000. The depreciation is:
\[
\text{Depreciation} = 52,800,000 - 45,800,000 = 7,000,000
\]

### Problem 2: Soybean Production vs. Fertilization (DAP)

#### a) Function \( P(x) \)

Here we are given two points:
- (46, 2.75) where 46 kg/ha DAP yields 2.75 tons/ha.
- (58, 3) where 58 kg/ha DAP yields 3 tons/ha.

We are to find the linear equation \( P(x) \) that relates production \( P(x) \) (tons/ha) to fertilization \( x \) (kg/ha). The slope is calculated as:
\[
m = \frac{3 - 2.75}{58 - 46} = \frac{0.25}{12} = \frac{1}{48}
\]

Now, use the point-slope form to find the equation:
\[
P(x) - 2.75 = \frac{1}{48}(x - 46)
\]
Simplify to:
\[
P(x) = \frac{1}{48}(x - 46) + 2.75
\]

This is the equation of \( P(x) \).

#### b) Function \( G(x) \)

For another variety of soybean, we are given:
- \( G(48) = 3.25 \)
- \( G(60) = 3.85 \)

Again, we calculate the slope:
\[
m = \frac{3.85 - 3.25}{60 - 48} = \frac{0.6}{12} = 0.05
\]

Now use the point-slope form with one of the points, say (48, 3.25):
\[
G(x) - 3.25 = 0.05(x - 48)
\]
Simplify:
\[
G(x) = 0.05(x - 48) + 3.25
\]

This is the equation of \( G(x) \).

---

Would you like more details on any part of this solution?

Here are some related questions to explore:
1. How would the depreciation function change if the machinery had a different annual depreciation rate?
2. What happens to the useful life of the machinery if its initial value is higher?
3. How would you modify the functions \( P(x) \) or \( G(x) \) if the relationship was not linear?
4. What are the implications of a longer useful life on the cost-effectiveness of the machinery?
5. How can we calculate the break-even point in terms of DAP application for soybean production?

**Tip:** Always check the units in problems involving both physical quantities (e.g., kg/ha, tons/ha) and monetary values to avoid confusion.

Solve problems on linear depreciation function and DAP fertilizer application for soybean production.

Solve linear depreciation problems for agricultural machinery and model soybean production with fertilizer applications. Learn key concepts such as depreciation equations, linear function slopes, and equation of a line. Suitable for students in Grades 10-12.

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