Math Problem Statement

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

Solution

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 AA be the starting point on the shore.
  • Let BB be the endpoint on the island.
  • Let CC be the point on the shore directly opposite BB, such that BCBC is perpendicular to the shore.
  • Let DD be the point on the shore where the pipeline changes from being onshore to underwater.
  • Distance AC=9AC = 9 km (distance along the shore from AA to the point directly opposite BB).
  • Distance BC=6BC = 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 xx be the distance from AA to DD (the point on the shore where the pipeline enters the water). So, 0x90 \leq x \leq 9.
  2. Onshore and Underwater Distances:

    • Onshore distance: AD=xAD = x.
    • Underwater distance: DB=(9x)2+62DB = \sqrt{(9 - x)^2 + 6^2} km (by the Pythagorean theorem).
  3. Cost Function:

    • Onshore cost: 4000×x4000 \times x.
    • Underwater cost: 5000×(9x)2+365000 \times \sqrt{(9 - x)^2 + 36}.
    • Total cost function C(x)C(x) is given by: C(x)=4000x+5000(9x)2+36C(x) = 4000x + 5000 \sqrt{(9-x)^2 + 36}

Minimizing the Cost Function:

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

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

  2. Set the Derivative to Zero: 4000=5000(9x)(9x)2+364000 = 5000 \cdot \frac{(9 - x)}{\sqrt{(9 - x)^2 + 36}} 45=9x(9x)2+36\frac{4}{5} = \frac{9 - x}{\sqrt{(9 - x)^2 + 36}}

  3. Solve for xx: 45(9x)2+36=9x\frac{4}{5} \sqrt{(9 - x)^2 + 36} = 9 - x Square both sides to eliminate the square root: (45)2[(9x)2+36]=(9x)2\left(\frac{4}{5}\right)^2 \left[(9 - x)^2 + 36\right] = (9 - x)^2 1625[(9x)2+36]=(9x)2\frac{16}{25} \left[(9 - x)^2 + 36\right] = (9 - x)^2 16(9x)2+576=25(9x)216(9 - x)^2 + 576 = 25(9 - x)^2 576=9(9x)2576 = 9(9 - x)^2 (9x)2=64(9 - x)^2 = 64 9x=8or9x=89 - x = 8 \quad \text{or} \quad 9 - x = -8 x=1(since x=17 is not possible within the range)x = 1 \quad \text{(since } x = 17 \text{ is not possible within the range)}

Conclusion:

The pipeline should be laid 11 km along the shore from AA to DD and then underwater from DD to BB 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 DD 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 ACAC or BCBC 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Pythagorean Theorem

Formulas

Cost Function: C(x) = 4000x + 5000 * sqrt((9 - x)^2 + 36)
Derivative of the Cost Function: C'(x) = 4000 + 5000 * (- (9 - x)) / sqrt((9 - x)^2 + 36)

Theorems

Pythagorean Theorem
Optimization using Calculus

Suitable Grade Level

Grades 11-12