A company wants to run a pipeline from a point A on the shore to a point B on an island which is 6km from the shore. It costs Birr 4000 per kilometre to run the pipe line on shore, and 5000 Birr per kilometre to run it underwater. There is a point     on the shore so that     is at the right angles to      . The straight shoreline is the line   . The distance     is 9km. Find how the pipeline should be laid to minimize the cost

To solve this problem, we need to determine the optimal path for laying the pipeline that minimizes the total cost. Let's break it down:

### Setup and Variables:

- Let \( A \) be the starting point on the shore.
- Let \( B \) be the endpoint on the island.
- Let \( C \) be the point on the shore directly opposite \( B \), such that \( BC \) is perpendicular to the shore.
- Let \( D \) be the point on the shore where the pipeline changes from being onshore to underwater.
- Distance \( AC = 9 \) km (distance along the shore from \( A \) to the point directly opposite \( B \)).
- Distance \( BC = 6 \) km (distance from the shore to the island).
- Cost of laying the pipeline onshore: 4000 Birr per km.
- Cost of laying the pipeline underwater: 5000 Birr per km.

### Finding the Minimum Cost Path:

1. **Define the Variables:**
   - Let \( x \) be the distance from \( A \) to \( D \) (the point on the shore where the pipeline enters the water). So, \( 0 \leq x \leq 9 \).

2. **Onshore and Underwater Distances:**
   - Onshore distance: \( AD = x \).
   - Underwater distance: \( DB = \sqrt{(9 - x)^2 + 6^2} \) km (by the Pythagorean theorem).

3. **Cost Function:**
   - Onshore cost: \( 4000 \times x \).
   - Underwater cost: \( 5000 \times \sqrt{(9 - x)^2 + 36} \).
   - Total cost function \( C(x) \) is given by:
     \[
     C(x) = 4000x + 5000 \sqrt{(9-x)^2 + 36}
     \]

### Minimizing the Cost Function:

To find the value of \( x \) that minimizes the cost, we need to take the derivative of \( C(x) \) and set it to zero.

1. **Derivative of the Cost Function:**
   \[
   C'(x) = 4000 + 5000 \cdot \frac{- (9 - x)}{\sqrt{(9 - x)^2 + 36}}
   \]

2. **Set the Derivative to Zero:**
   \[
   4000 = 5000 \cdot \frac{(9 - x)}{\sqrt{(9 - x)^2 + 36}}
   \]
   \[
   \frac{4}{5} = \frac{9 - x}{\sqrt{(9 - x)^2 + 36}}
   \]

3. **Solve for \( x \):**
   \[
   \frac{4}{5} \sqrt{(9 - x)^2 + 36} = 9 - x
   \]
   Square both sides to eliminate the square root:
   \[
   \left(\frac{4}{5}\right)^2 \left[(9 - x)^2 + 36\right] = (9 - x)^2
   \]
   \[
   \frac{16}{25} \left[(9 - x)^2 + 36\right] = (9 - x)^2
   \]
   \[
   16(9 - x)^2 + 576 = 25(9 - x)^2
   \]
   \[
   576 = 9(9 - x)^2
   \]
   \[
   (9 - x)^2 = 64
   \]
   \[
   9 - x = 8 \quad \text{or} \quad 9 - x = -8
   \]
   \[
   x = 1 \quad \text{(since } x = 17 \text{ is not possible within the range)}
   \]

### Conclusion:

The pipeline should be laid \( 1 \) km along the shore from \( A \) to \( D \) and then underwater from \( D \) to \( B \) to minimize the cost.

### Follow-Up Questions:

1. How would the cost change if the cost of laying the underwater pipeline increased?
2. What would be the optimal point \( D \) if the cost of laying both onshore and underwater pipelines were the same?
3. How does the total cost compare when laying the entire pipeline underwater versus the optimal solution?
4. What is the cost difference between laying the pipeline onshore versus underwater?
5. How would the solution change if the distance \( AC \) or \( BC \) were different?

### Tip:

When solving optimization problems involving cost and distance, always start by defining variables, expressing the cost function in terms of these variables, and then use calculus (derivatives) to find the minimum or maximum values.

A company wants to run a pipeline from a point A on the shore to a point B on an island which is 6km from the shore. It costs Birr 4000 per kilometre to run the pipeline on shore, and 5000 Birr per kilometre to run it underwater. There is a point on the shore so that is at the right angles to . The straight shoreline is the line . The distance is 9km. Find how the pipeline should be laid to minimize the cost.

Discover how to optimize the cost of laying a pipeline between a shore and an island using calculus and the Pythagorean theorem. This problem involves deriving a cost function and finding its minimum value, suitable for students in Grades 11-12.

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