13.3 Closed interval method

The Extreme Value Theorem is a game-changer for finding the highest and lowest points of continuous functions. It guarantees these points exist on closed intervals, making it possible to solve real-world problems like maximizing profit or optimizing designs.

The closed interval method is a powerful tool that puts the Extreme Value Theorem into action. By checking critical points and endpoints, we can pinpoint exactly where a function reaches its peaks and valleys, giving us crucial insights for practical applications.

The Closed Interval Method and the Extreme Value Theorem

Extreme Value Theorem implications

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• States continuous function $f$ on closed interval $[a, b]$ attains absolute maximum value $f(c)$ and absolute minimum value $f(d)$ at some numbers $c$ and $d$ in $[a, b]$
• Guarantees existence of absolute extrema for continuous functions on closed intervals (sine function)
• Allows use of closed interval method to find absolute extrema
• Provides foundation for solving optimization problems involving continuous functions on closed intervals (maximizing profit)

Steps of closed interval method

1. Ensure function $f$ is continuous on closed interval $[a, b]$
2. Find all critical numbers of $f$ in interval $[a, b]$
   • Critical numbers are values of $x$ where $f'(x) = 0$ or $f'(x)$ does not exist (cusps, discontinuities)
3. Evaluate $f$ at each critical number found in Step 2
4. Evaluate $f$ at endpoints of interval, $a$ and $b$
5. Compare all function values obtained in Steps 3 and 4
   • Largest value is absolute maximum (global maximum)
   • Smallest value is absolute minimum (global minimum)

Absolute extrema on closed intervals

• Given continuous function $f$ on closed interval $[a, b]$, follow steps of closed interval method
  • Find all critical numbers of $f$ in $[a, b]$
  • Evaluate $f$ at each critical number and endpoints $a$ and $b$
  • Compare function values to determine absolute maximum and minimum
• Example: Find absolute extrema of $f(x) = x^3 - 3x^2 - 9x + 1$ on interval $[-2, 4]$
  • $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$
  • Critical numbers: $x = -1, 3$
  • Evaluate $f$ at critical numbers and endpoints: $f(-2), f(-1), f(3), f(4)$
  • Compare values to find absolute maximum and minimum

Optimization with closed interval method

• Optimization problems involve finding maximum or minimum value of function subject to constraints
• Steps to solve optimization problems using closed interval method
  1. Identify objective function (function to be maximized or minimized)
  2. Determine constraints on variables (domain restrictions)
  3. Express objective function in terms of single variable
  4. Find closed interval over which objective function is defined
  5. Apply closed interval method to find absolute extrema of objective function on closed interval
  6. Interpret results in context of original problem
• Example: Rectangular garden has perimeter of 200 ft. Find dimensions that maximize area.
  • Objective function: $A(x) = x(100 - x)$, where $x$ is width and $100 - x$ is length
  • Constraints: $0 < x < 100$
  • Closed interval: $[0, 100]$
  • Apply closed interval method to find maximum area