Local Maxima & Minima
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Local Maximum
f(c) is a local maximum if f(c) ≥ f(x) for all x near c.
The function rises to c, then falls.
The function rises to c, then falls.
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Local Minimum
f(c) is a local minimum if f(c) ≤ f(x) for all x near c.
The function falls to c, then rises.
The function falls to c, then rises.
Critical Points
A point x = c is called a critical point (or stationary point) if f'(c) = 0 or f'(c) does not exist. Local maxima and minima can only occur at critical points.
Three Methods to Find Maxima/Minima
1st Derivative Test
- Find f'(x) and solve f'(x) = 0 → get c.
- Check sign of f'(x) near c:
- + → – : Local Max
- – → + : Local Min
- No change: Neither
2nd Derivative Test
- Find f'(x) = 0 → get c.
- Find f''(c):
- f''(c) < 0 : Local Max
- f''(c) > 0 : Local Min
- f''(c) = 0 : Test fails → use 1st derivative test
Absolute (Global) Maxima & Minima
On a Closed Interval [a, b]
To find the absolute maximum and minimum of f on [a, b]:
1. Find all critical points c in (a, b) where f'(c) = 0.
2. Calculate f(a), f(b), and f(c) for each critical point.
3. The largest value is the absolute maximum; the smallest is the absolute minimum.
Point of Inflection: If f'(c) = 0 but f''(c) = 0 too, it could be an inflection point where the curve changes concavity. Use the 1st derivative test to confirm.
Key Formulae
- 1For rectangle with perimeter P: Area is maximum when it is a square — all sides equal.
- 2Cylinder of fixed volume: Surface area is minimum when height = diameter (h = 2r).
- 3Sum of two numbers = S: Product is maximum when both numbers equal S/2.
- 4Open box from square sheet: Volume is maximum at a specific cut corner x = a/6 (if sheet is a × a).
Solved Examples
Example 01
Local max/min of f(x) = x³ – 3x² + 4
Q. Find local maxima and minima of f(x) = x³ – 3x² + 4 using 2nd derivative test.
f'(x) = 3x² – 6x = 3x(x – 2) = 0 → x = 0 or x = 2
f''(x) = 6x – 6
At x=0: f''(0) = –6 < 0 → Local Maximum. f(0) = 4
At x=2: f''(2) = 6 > 0 → Local Minimum. f(2) = 8–12+4 = 0
Local Max = 4 at x = 0 | Local Min = 0 at x = 2
Example 02
Absolute max/min on closed interval
Q. Find absolute maxima and minima of f(x) = x³ – 3x on [–2, 3].
f'(x) = 3x² – 3 = 0 → x² = 1 → x = ±1 (both in [–2,3])
f(–2) = –8+6 = –2; f(–1) = –1+3 = 2; f(1) = 1–3 = –2; f(3) = 27–9 = 18
Compare: –2, 2, –2, 18
Absolute Maximum = 18 at x = 3 | Absolute Minimum = –2 at x = –2 and x = 1
Example 03
Optimization — Rectangle with maximum area
Q. A farmer has 80 m of fencing. Find dimensions of a rectangular field with maximum area.
Let length = x, width = y. Perimeter: 2(x + y) = 80 → x + y = 40 → y = 40 – x
Area A = x·y = x(40 – x) = 40x – x²
dA/dx = 40 – 2x = 0 → x = 20
d²A/dx² = –2 < 0 → Maximum at x = 20
y = 40 – 20 = 20. Area = 20 × 20 = 400 m²
∴ Dimensions: 20 m × 20 m (square gives maximum area = 400 m²)