Find an optimum, then justify it.
The first derivative describes slope; the second describes how slope changes. A positive second derivative means concave up, and a negative one means concave down. An inflection needs a change in concavity, so f″ = 0 is only a candidate test.
For optimisation, define a feasible domain, use constraints to reduce the model to one variable, then find stationary inputs. Check endpoints and classify the candidates. A mathematically valid stationary input may be outside the physical domain or may minimise when the question asks for a maximum.
Try it. Watch it change.
The graph is calculated from the displayed rule. Drag the highlighted point or use the sliders.
Explore: Display the cubic and its first and second derivatives. Compare stationary points with the inflection point.
A rectangle has perimeter 24 m. Find the maximum area.
- If one side is x, the other is 12 − x, with 0 < x < 12.
- A = x(12 − x); A′ = 12 − = 0 gives x = 6.
- A″ = −2 < 0, so the maximum is 36 , achieved by a square.
Assumed knowledge
Differentiation, stationary points and modelling constraints.
Learning checkpoints & source
YOUR LEARNING CHECKPOINT- Use first and second derivatives to classify features.
- Build and check an optimisation model.