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 AA to BB, 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 0.750.75 miles wide.
  • AA and BB are on opposite banks, with BB located 22 miles downstream from AA.
  • Cost per mile in water: $2million\$2 \, \text{million}.
  • Cost per mile on land: $1million\$1 \, \text{million}.

Approach

  1. Variables: Let xx represent the distance downstream from AA where the pipeline crosses the river (the point where the pipeline exits the water).

  2. Land Distance: The length of the pipeline on land is 2x2 - x.

  3. Water Distance: The length of the pipeline in water is x2+0.752\sqrt{x^2 + 0.75^2} (using the Pythagorean theorem).

  4. Cost Function: The total cost C(x)C(x) in millions of dollars is: C(x)=2x2+0.752+(2x)C(x) = 2 \sqrt{x^2 + 0.75^2} + (2 - x) where 2x2+0.7522 \sqrt{x^2 + 0.75^2} is the water cost, and (2x)(2 - x) is the land cost.

  5. Optimize: We minimize C(x)C(x) by finding its derivative and solving dCdx=0\frac{dC}{dx} = 0.


Steps

  1. Derivative of C(x)C(x): dCdx=4xx2+0.7521\frac{dC}{dx} = \frac{4x}{\sqrt{x^2 + 0.75^2}} - 1

  2. Solve dCdx=0\frac{dC}{dx} = 0: 4xx2+0.752=1\frac{4x}{\sqrt{x^2 + 0.75^2}} = 1 4x=x2+0.7524x = \sqrt{x^2 + 0.75^2} Square both sides: 16x2=x2+0.75216x^2 = x^2 + 0.75^2 15x2=0.562515x^2 = 0.5625 x2=0.562515x^2 = \frac{0.5625}{15} x2=0.0375,x=0.03750.1938miles.x^2 = 0.0375, \quad x = \sqrt{0.0375} \approx 0.1938 \, \text{miles}.

  3. Compute C(x)C(x):

    • Water distance: 0.19382+0.7520.7755miles.\sqrt{0.1938^2 + 0.75^2} \approx 0.7755 \, \text{miles}.
    • Land distance: 20.1938=1.8062miles.2 - 0.1938 = 1.8062 \, \text{miles}.
    • Cost: C(x)=2(0.7755)+1(1.8062)=1.551+1.80623.3572million dollars.C(x) = 2(0.7755) + 1(1.8062) = 1.551 + 1.8062 \approx 3.3572 \, \text{million dollars}.

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:

  1. What happens if the river width changes?
  2. How would the cost change if the water cost doubles?
  3. Can we find the crossing point graphically instead of using calculus?
  4. What if BB were further downstream?
  5. 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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