Stationary points and optimisation

At a turning point the gradient is zero. Setting dy/dx = 0 finds maximum and minimum points, which solves optimisation problems: the biggest area, the cheapest cost, the shortest time.

Key facts

  • Stationary points: solve dy/dx = 0.
  • Second derivative: positive means minimum, negative means maximum.
  • Optimisation: write the quantity in terms of one variable, differentiate, set to zero.
  • Increasing where dy/dx > 0; decreasing where dy/dx < 0.

Worked examples

Example 1

Find and classify the stationary points of y = x³ − 3x.

  1. dy/dx = 3x² − 3 = 0 gives x = ±1.
  2. d²y/dx² = 6x: at x = 1 it’s 6 (minimum), at x = −1 it’s −6 (maximum).
  3. Points: (1, −2) minimum and (−1, 2) maximum.

Answer: Max (−1, 2), min (1, −2)

Example 2

A farmer has 40 m of fencing to make a rectangular pen against a wall (3 sides). Find the largest area.

  1. Width x, length 40 − 2x: A = x(40 − 2x) = 40x − 2x².
  2. dA/dx = 40 − 4x = 0 x = 10
  3. A = 10 × 20 = 200 m²

Answer: 200 m²