Math Problem Statement

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

Solution

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: 22 km.
  2. Distance Along the Bank: 66 km.
  3. Location of Point PP Across the River: Let xx be the distance from the refinery to the point directly opposite PP on the north bank. xx lies between 00 and 66.

Cost Function

The pipeline is divided into two sections:

  1. Under the River: From the refinery to point PP, which forms the hypotenuse of a right triangle with legs xx and 22. The distance is x2+22\sqrt{x^2 + 2^2}.
  2. Over Land: From PP to the storage tanks. The distance is 6x6 - x.

The costs are:

  • 800,000800,000 per km under the river.
  • 400,000400,000 per km over land.

Total Cost

Cost(x)=800,000x2+4+400,000(6x)\text{Cost}(x) = 800,000 \cdot \sqrt{x^2 + 4} + 400,000 \cdot (6 - x)

Minimization

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

Derivative of the Cost Function

Cost(x)=800,000x2+4+400,000(6x)\text{Cost}(x) = 800,000 \cdot \sqrt{x^2 + 4} + 400,000 \cdot (6 - x)

  1. Compute the derivative: ddxCost(x)=800,000xx2+4400,000\frac{d}{dx} \text{Cost}(x) = 800,000 \cdot \frac{x}{\sqrt{x^2 + 4}} - 400,000

  2. Set the derivative to 00: 800,000xx2+4=400,000800,000 \cdot \frac{x}{\sqrt{x^2 + 4}} = 400,000

  3. Simplify: xx2+4=12\frac{x}{\sqrt{x^2 + 4}} = \frac{1}{2}

  4. Solve for xx: x2=43x^2 = \frac{4}{3} x=2331.155km.x = \frac{2\sqrt{3}}{3} \approx 1.155 \, \text{km}.

Thus, PP should be located about 1.1551.155 km east of the refinery on the north bank.

Verification and Conclusion

To verify, substitute x=1.155x = 1.155 into the cost function to ensure it is minimized. The result confirms that placing PP at this location minimizes the construction cost.


Would you like a detailed step-by-step verification or graph for better visualization?

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.

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Math Problem Analysis

Mathematical Concepts

Optimization
Derivatives
Cost Minimization

Formulas

Cost(x) = 800,000 * sqrt(x^2 + 4) + 400,000 * (6 - x)
Derivative: d/dx Cost(x) = 800,000 * (x / sqrt(x^2 + 4)) - 400,000

Theorems

Optimization using derivatives

Suitable Grade Level

Grades 11-12 (Advanced Calculus/Optimization)