Maxima & Minima

1. Set the first derivative to zero to locate the critical points.
2. Evaluate the second derivative there: negative means a maximum, positive means a minimum.
3. Check the endpoints of the interval separately.

Worked example

A haul road is being regraded. Along the 500 m stretch in question, a survey fits the existing ground to y = x³ − 6x² + 9x + 2, where x is distance along the road in hundreds of metres (0 ≤ x ≤ 5) and y is elevation in metres above a site datum. The road is 8 m wide. Where on this stretch is the ground highest?

1. y′ = 3x² − 12x + 9. Setting it to zero gives x = 1 and x = 3.
2. y″ = 6x − 12. It is −6 at x = 1: a local maximum. It is 6 at x = 3: a local minimum.
3. Now evaluate the ends of the interval, x = 0 and x = 5, which the derivative test cannot find.