Math Problem Statement
A construction company is given a contract to build a pipeline from point A on the bank of the river to point B on the opposite bank and 2 miles downstream. If the river is 0.75 miles wide, what is the cheapest construction cost to build the pipeline if it costs $1 millions per mile for a pipeline built on dry land and $2 million for each mile in water?
Solution
To minimize the construction cost of the pipeline from to , we need to determine the optimal location to cross the river, minimizing the total cost of both the water and land segments of the pipeline.
Problem Setup
- The river is miles wide.
- and are on opposite banks, with located miles downstream from .
- Cost per mile in water: .
- Cost per mile on land: .
Approach
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Variables: Let represent the distance downstream from where the pipeline crosses the river (the point where the pipeline exits the water).
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Land Distance: The length of the pipeline on land is .
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Water Distance: The length of the pipeline in water is (using the Pythagorean theorem).
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Cost Function: The total cost in millions of dollars is: where is the water cost, and is the land cost.
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Optimize: We minimize by finding its derivative and solving .
Steps
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Derivative of :
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Solve : Square both sides:
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Compute :
- Water distance:
- Land distance:
- Cost:
Final Answer
The cheapest construction cost is approximately $3.36 million.
Would you like further clarification or details about the process?
Here are 5 related questions you might find interesting:
- What happens if the river width changes?
- How would the cost change if the water cost doubles?
- Can we find the crossing point graphically instead of using calculus?
- What if were further downstream?
- How does the total cost change if both land and water costs are equal?
Tip: Drawing a diagram of the situation often helps clarify problems involving distances and geometry.
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Math Problem Analysis
Mathematical Concepts
Optimization
Geometry
Calculus
Formulas
C(x) = 2 * sqrt(x^2 + 0.75^2) + (2 - x)
Theorems
Pythagorean theorem
Optimization using derivatives
Suitable Grade Level
Grades 11-12