Learn & Review: Optimization Problem in Calculus

Summary of Optimization Problems in Calculus

This content introduces optimization problems in calculus using two real-life examples: maximizing the area of a rectangle made from a wire and minimizing the travel time to an island. The core idea is to use calculus to find the exact dimensions or path that result in a maximum or minimum value.

1. Maximizing Rectangle Area

• Problem: Given a wire of a fixed length (e.g., 100 meters), what dimensions of a rectangle formed from it will yield the maximum possible area?
• Examples:
  • Length 40m, Width 10m = Area 400 sq units
  • Length 35m, Width 15m = Area 525 sq units
  • Square (Length 25m, Width 25m) = Area 625 sq units
• Concept: This is an optimization problem where the goal is to maximize the area. Calculus provides a method to find the exact dimensions, rather than guessing.

2. Minimizing Travel Time to an Island

• Scenario: Alex is at a seaside cabin located 6 km west of a point on the beach. An island is 2 km directly north of that same point on the beach. Alex can run along the beach at 8 km/h and swim at 3 km/h.
• Objective: Find the path that allows Alex to reach the island in the shortest possible time.
• Possible Paths:
  • Swim directly to the island.
  • Run to the point on the beach closest to the island and then swim.
  • Run part of the way along the beach and then swim.
• Optimization Goal: Minimize the total travel time.

Mathematical Formulation

• Variables:
  • Let x be the distance Alex runs along the beach (in km).
  • Let y be the distance Alex swims (in km).
• Speeds:
  • Running speed = 8 km/h
  • Swimming speed = 3 km/h
• Time Calculation:
  • Time = Distance / Speed
  • Time spent running = x / 8
  • Time spent swimming = y / 3
  • Total time t = (x / 8) + (y / 3)
• Relating x and y:
  • The distance along the beach from Alex's cabin to the point where he starts swimming is x.
  • The remaining distance along the beach to the point directly opposite the island is 6 - x.
  • The distance from the island to the beach is 2 km.
  • Using the Pythagorean theorem, y² = 2² + (6 - x)².
  • Therefore, y = sqrt(2² + (6 - x)²).
• Total Time as a Function of x: By substituting the expression for y into the total time equation, t becomes a function of a single variable, x: t(x) = (x / 8) + (sqrt(4 + (6 - x)²)) / 3

Using Calculus for Optimization

• Concept: To find the minimum travel time, we need to find the value of x that minimizes the function t(x).
• Derivative: The derivative of t(x) with respect to x (dt/dx) represents the rate of change of time as x changes. It is also the slope of the t(x) graph.
• Finding the Minimum: The minimum point on the graph of t(x) occurs where the slope is zero.
  • Set the derivative dt/dx equal to zero.
  • Solve the resulting equation for x.
• Solution:
  • The derivative of t(x) is calculated (steps not shown in detail).
  • Setting dt/dx = 0 and solving for x yields: x = 6 - 6 / sqrt(55)
  • This value is approximately 5.19 km.
• Conclusion: Alex should run approximately 5.19 km along the beach before swimming to the island to achieve the fastest possible travel time.

Key Takeaway

Optimization problems in calculus involve:

1. Formulating a function to represent the quantity to be maximized or minimized (e.g., area, time).
2. Expressing this function in terms of a single variable.
3. Using the derivative of the function.
4. Setting the derivative to zero and solving for the variable to find the point of maximum or minimum.