An oil refinery is located on the north bank of a straight river that is 2km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. The cost of laying pipe is 800,000/km under the river to the tanks and 400,000/km over land. To minimize the cost where should P be located

To solve this optimization problem, we need to minimize the cost of constructing the pipeline. Let's calculate step by step:

### Variables
1. **River Width**: \( 2 \) km.
2. **Distance Along the Bank**: \( 6 \) km.
3. **Location of Point \( P \) Across the River**: Let \( x \) be the distance from the refinery to the point directly opposite \( P \) on the north bank. \( x \) lies between \( 0 \) and \( 6 \).

### Cost Function
The pipeline is divided into two sections:
1. **Under the River**: From the refinery to point \( P \), which forms the hypotenuse of a right triangle with legs \( x \) and \( 2 \). The distance is \( \sqrt{x^2 + 2^2} \).
2. **Over Land**: From \( P \) to the storage tanks. The distance is \( 6 - x \).

The costs are:
- \( 800,000 \) per km under the river.
- \( 400,000 \) per km over land.

#### Total Cost
\[
\text{Cost}(x) = 800,000 \cdot \sqrt{x^2 + 4} + 400,000 \cdot (6 - x)
\]

### Minimization
We minimize \( \text{Cost}(x) \) by finding its derivative and solving for \( \frac{d}{dx} \text{Cost}(x) = 0 \).

#### Derivative of the Cost Function
\[
\text{Cost}(x) = 800,000 \cdot \sqrt{x^2 + 4} + 400,000 \cdot (6 - x)
\]

1. Compute the derivative:
\[
\frac{d}{dx} \text{Cost}(x) = 800,000 \cdot \frac{x}{\sqrt{x^2 + 4}} - 400,000
\]

2. Set the derivative to \( 0 \):
\[
800,000 \cdot \frac{x}{\sqrt{x^2 + 4}} = 400,000
\]

3. Simplify:
\[
\frac{x}{\sqrt{x^2 + 4}} = \frac{1}{2}
\]

4. Solve for \( x \):
\[
x^2 = \frac{4}{3}
\]
\[
x = \frac{2\sqrt{3}}{3} \approx 1.155 \, \text{km}.
\]

Thus, \( P \) should be located about \( 1.155 \) km east of the refinery on the north bank.

### Verification and Conclusion
To verify, substitute \( x = 1.155 \) into the cost function to ensure it is minimized. The result confirms that placing \( P \) at this location minimizes the construction cost.

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### Related Questions:
1. What is the total cost of constructing the pipeline at the optimal location?
2. How does the cost change if the cost per km under the river increases to 1,000,000?
3. What happens if the width of the river increases to 3 km?
4. Can we generalize this solution to other values of river width and tank distance?
5. What are the potential real-world constraints that might alter this solution?

### Tip:
When optimizing functions, always check boundary values to ensure the calculated solution is indeed the minimum.

An oil refinery is located on the north bank of a straight river that is 2km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. The cost of laying pipe is 800,000/km under the river to the tanks and 400,000/km over land. To minimize the cost where should P be located?

Discover how to minimize the cost of constructing a pipeline across a 2km wide river with a 6km eastward storage location. Learn to apply optimization techniques using derivatives and cost functions for advanced calculus students. Ideal for Grades 11-12.

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