Math Problem Statement

A rancher travels in his truck from his ranch to the nearest city (see diagram). He can travel across the open range of his ranch at a speed of 40 mph and along the highway at 50 mph. Determine the route he should take in order to reach the city in the shortest amount of time. Your supporting work must include all computations using the proper optimization method. Decimal answers are acceptable.

Solution

This optimization problem requires determining the shortest time for the rancher to travel from his ranch to the city using two distinct travel speeds: 40 mph across the open range and 50 mph along the highway.

Let's break this down step by step:

Variables and Equations:

  1. Diagram Information:

    • Distance across the range: Hypotenuse of a triangle with legs of 20 miles (north-south) and xx miles (east-west).
    • Distance along the highway: 60x60 - x, where xx is the portion of the eastward highway covered by traveling across the range.
  2. Travel Speeds:

    • Across the range: 40 mph
    • Along the highway: 50 mph
  3. Objective: Minimize total travel time TT, which is the sum of the time taken across the range and along the highway.

  4. Distance Across the Range: The hypotenuse cc is given by: c=202+x2=400+x2c = \sqrt{20^2 + x^2} = \sqrt{400 + x^2}

  5. Time Across the Range: Time=DistanceRate=400+x240\text{Time} = \frac{\text{Distance}}{\text{Rate}} = \frac{\sqrt{400 + x^2}}{40}

  6. Time Along the Highway: Time=DistanceRate=60x50\text{Time} = \frac{\text{Distance}}{\text{Rate}} = \frac{60 - x}{50}

  7. Total Time: T(x)=400+x240+60x50T(x) = \frac{\sqrt{400 + x^2}}{40} + \frac{60 - x}{50}

Optimization:

  1. Minimize T(x)T(x): Take the derivative of T(x)T(x) and solve for xx where T(x)=0T'(x) = 0.

    T(x)=140x400+x2150T'(x) = \frac{1}{40} \cdot \frac{x}{\sqrt{400 + x^2}} - \frac{1}{50}

    Simplify: T(x)=x40400+x2150T'(x) = \frac{x}{40\sqrt{400 + x^2}} - \frac{1}{50}

  2. Set T(x)=0T'(x) = 0: x40400+x2=150\frac{x}{40\sqrt{400 + x^2}} = \frac{1}{50}

    Multiply through by 40400+x25040\sqrt{400 + x^2} \cdot 50: 50x=40400+x250x = 40\sqrt{400 + x^2}

    Square both sides: 2500x2=1600(400+x2)2500x^2 = 1600(400 + x^2)

    Expand: 2500x2=640,000+1600x22500x^2 = 640,000 + 1600x^2

    Solve for x2x^2: 900x2=640,000900x^2 = 640,000

    x2=640,000900=711.11x^2 = \frac{640,000}{900} = 711.11

    x=711.1126.68x = \sqrt{711.11} \approx 26.68

  3. Interpretation:

    • The rancher should travel approximately 26.68 miles east across the open range before switching to the highway.

Total Time:

Substitute x=26.68x = 26.68 into T(x)T(x) to find the minimum travel time: T(x)=400+(26.68)240+6026.6850T(x) = \frac{\sqrt{400 + (26.68)^2}}{40} + \frac{60 - 26.68}{50}

This completes the solution process. Let me know if you’d like further explanation or calculations!


Related Questions:

  1. How does the travel time vary if the highway speed increases to 60 mph?
  2. What happens if the rancher takes a direct diagonal path to the city?
  3. How would the calculation change if the open-range speed decreases to 30 mph?
  4. What optimization techniques could be used to confirm the solution?
  5. Could this problem be solved graphically instead of algebraically?

Tip:

Always visualize the geometry of optimization problems to better understand the variables and constraints involved.

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Math Problem Analysis

Mathematical Concepts

Optimization
Geometry
Calculus

Formulas

Distance = Speed × Time
Hypotenuse: c = √(a² + b²)
Derivative for critical points: T'(x) = 0

Theorems

Pythagorean Theorem
First Derivative Test

Suitable Grade Level

Grades 11-12 or College Level