Overview

This topic applies derivatives to real-world problems: finding maximum or minimum values in optimization, and solving problems where two or more quantities change over time using related rates. Both require modeling with functions and applying differentiation systematically.

Key Concepts and Structures

• Optimization:
  • Find a quantity to maximize or minimize (e.g., area, cost, profit)
  • Write an equation for the quantity to optimize (objective function)
  • Use constraints to reduce to a single-variable function
  • Take derivative, find critical points, test for max/min
• Related Rates:
  • Two or more variables change over time (usually with respect to t)
  • Differentiate both sides of an equation using the chain rule: \frac{d}{dt}
  • Plug in known values and solve for the unknown rate
• Common Shapes: Area of rectangle A = lw, volume of cone V = (1/3)\pi r^2h, Pythagorean Theorem a^2 + b^2 = c^2

Quick Tip

Practice Problems

1. A rectangle has a perimeter of 20 units. What is the maximum possible area?
   Show Solution

   Step 1: Let width = w, length = l. Perimeter: 2l + 2w = 20

   Simplify: l = 10 - w

   Area: A = lw = w(10 - w)

   Take derivative: A' = 10 - 2w. Set A' = 0 → w = 5

   Final Answer: Maximum area is at w = l = 5 → A = 25

2. A balloon rises at 4 ft/sec. A person stands 10 ft away. How fast is the distance from the person to the balloon changing when the balloon is 24 ft high?
   Show Solution

   Step 1: Use Pythagorean Theorem: z^2 = x^2 + y^2, where z is the distance

   x = 10 (constant), y = height, \frac{dy}{dt} = 4

   Differentiating: 2z \cdot \frac{dz}{dt} = 2y \cdot \frac{dy}{dt}

   At y = 24: z = \sqrt{10^2 + 24^2} = 26

   Solve: 2(26) \cdot \frac{dz}{dt} = 2(24)(4) → \frac{dz}{dt} = \frac{192}{52} = \frac{48}{13}

   Final Answer: Distance increasing at \frac{48}{13} ft/sec