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Maxima & Minima

Short Notes

Maxima and Minima (extrema) are the largest and smallest values of a function. Finding extrema is crucial for optimization problems.

  • Local Maximum: f(c) ≥ f(x) for all x near c
  • Local Minimum: f(c) ≤ f(x) for all x near c
  • Global (Absolute) Maximum/Minimum: Largest/smallest value over the entire domain
  • Critical Point: Where f'(x) = 0 or f'(x) doesn't exist

Key Theories & Formulas

1. First Derivative Test

At a critical point c where f'(c) = 0: | Sign Change of f'(x) | Conclusion | |--------------------|------------| | + to − (increasing to decreasing) | Local Maximum | | − to + (decreasing to increasing) | Local Minimum | | No sign change | Neither (inflection point) |

2. Second Derivative Test

At a critical point c where f'(c) = 0: | f''(c) | Conclusion | |--------|------------| | f''(c) < 0 | Local Maximum (concave down) | | f''(c) > 0 | Local Minimum (concave up) | | f''(c) = 0 | Inconclusive (use first derivative test) |

3. Global Extrema on Closed Interval [a, b]

  1. Find all critical points in (a, b)
  2. Evaluate f at critical points and endpoints
  3. Compare: largest = global max, smallest = global min

4. Multivariable Optimization (Two Variables)

For f(x, y), find critical points where \(f_x = 0\) and \(f_y = 0\).

Second Derivative Test using Hessian determinant: $\(D = f_{xx} \cdot f_{yy} - (f_{xy})^2\)$

Condition Conclusion
D > 0 and \(f_{xx}\) < 0 Local Maximum
D > 0 and \(f_{xx}\) > 0 Local Minimum
D < 0 Saddle Point
D = 0 Inconclusive

5. Constrained Optimization (Lagrange Multipliers)

To optimize f(x, y) subject to constraint g(x, y) = 0: $\(\nabla f = \lambda \nabla g\)$

Solve: \(f_x = \lambda g_x\), \(f_y = \lambda g_y\), \(g(x, y) = 0\)

Example Problems

Problem 1: Find local extrema of f(x) = x³ - 3x + 2

Solution:

  1. f'(x) = 3x² - 3 = 0 ⟹ x = ±1
  2. f''(x) = 6x
  3. At x = 1: f''(1) = 6 > 0 → Local Minimum, f(1) = 0
  4. At x = -1: f''(-1) = -6 < 0 → Local Maximum, f(-1) = 4

Problem 2: Find global extrema of f(x) = x³ - 3x on [-2, 2]

Solution:

  1. Critical points: f'(x) = 3x² - 3 = 0 ⟹ x = ±1
  2. Evaluate:
  3. f(-2) = -8 + 6 = -2
  4. f(-1) = -1 + 3 = 2
  5. f(1) = 1 - 3 = -2
  6. f(2) = 8 - 6 = 2
  7. Global Max = 2 at x = -1 and x = 2
  8. Global Min = -2 at x = -2 and x = 1

Problem 3: Classify critical points of f(x, y) = x² + y² - 2x - 4y + 5

Solution:

  1. \(f_x = 2x - 2 = 0\) ⟹ x = 1
  2. \(f_y = 2y - 4 = 0\) ⟹ y = 2
  3. Critical point: (1, 2)
  4. \(f_{xx} = 2\), \(f_{yy} = 2\), \(f_{xy} = 0\)
  5. D = 2 × 2 - 0 = 4 > 0 and \(f_{xx}\) > 0
  6. Local Minimum at (1, 2), f(1, 2) = 1 + 4 - 2 - 8 + 5 = 0

Hardest GATE Questions

Topic: Second Derivative Test Failure

Question (GATE 2019 Style): Classify the critical point of f(x) = x⁴ at x = 0.

Solution:

  1. f'(x) = 4x³ = 0 ⟹ x = 0
  2. f''(x) = 12x², f''(0) = 0 → Inconclusive!
  3. Use first derivative test:
  4. For x < 0: f'(x) = 4x³ < 0 (decreasing)
  5. For x > 0: f'(x) = 4x³ > 0 (increasing)
  6. Sign changes from − to + → Local Minimum
  7. Alternative: f(x) = x⁴ ≥ 0 and f(0) = 0 → Global Minimum

Why it's hard: Second derivative test fails; must use first derivative test.

Topic: Multivariable Saddle Point

Question (GATE 2017 Variant): Classify all critical points of f(x, y) = x³ - 3xy + y³

Solution:

  1. \(f_x = 3x² - 3y = 0\) ⟹ y = x²
  2. \(f_y = -3x + 3y² = 0\) ⟹ x = y²
  3. Substitute: x = (x²)² = x⁴ ⟹ x⁴ - x = 0 ⟹ x(x³ - 1) = 0
  4. x = 0 or x = 1
  5. Critical points: (0, 0) and (1, 1)
  6. Hessian analysis:
  7. \(f_{xx} = 6x\), \(f_{yy} = 6y\), \(f_{xy} = -3\)
  8. At (0, 0): D = 0 - 9 = -9 < 0 → Saddle Point
  9. At (1, 1): D = 36 - 9 = 27 > 0, \(f_{xx}\) = 6 > 0 → Local Minimum

Topic: Lagrange Multipliers

Question: Find the maximum value of f(x, y) = xy subject to x² + y² = 8.

Solution:

  1. \(\nabla f = (y, x)\), \(\nabla g = (2x, 2y)\) where g = x² + y² - 8
  2. \(y = 2\lambda x\) and \(x = 2\lambda y\)
  3. From these: \(y = 2\lambda(2\lambda y) = 4\lambda² y\)
  4. Either y = 0 or λ² = 1/4 ⟹ λ = ±1/2
  5. If λ = 1/2: y = x, so 2x² = 8 ⟹ x = ±2
  6. Points: (2, 2), (-2, -2), (2, -2), (-2, 2)
  7. f(2, 2) = 4, f(-2, -2) = 4, f(2, -2) = -4, f(-2, 2) = -4
  8. Maximum = 4 at (2, 2) and (-2, -2)

References