Two ways to test a stationary point, optimisation in context, and points of inflexion where the gradient is nowhere near zero.
A closed cylindrical can must hold 250π cm³. Make it tall and thin and you use a lot of metal in the sides; make it short and wide and you use a lot in the lid and base. Move the radius and find the cheapest shape.
The minimum is not where it looks obvious. Calculus finds it exactly.
Two methods, and either is acceptable:
Justify that it is a minimum. d²S/dr² = 4π + 1000π/r³, which is positive for every positive r, so it is a minimum everywhere it can be. One line, and usually a mark.
At a point of inflexion f″ = 0 and changes sign. The gradient there may be zero, and very often is not.
| f′ there | f″ there | What it is | |
|---|---|---|---|
| y = x³ − 6x² + 9x at x = 2 | −3 | 0, changes sign | inflexion, non-stationary |
| y = x³ at x = 0 | 0 | 0, changes sign | inflexion, stationary |
| y = x⁴ at x = 0 | 0 | 0, no sign change | not an inflexion: a minimum |
f″ = 0 is not enough. For y = x⁴, f″ = 12x² is zero at the origin and positive on both sides, so the curve never stops being concave up. Always check the sign changes, the same way you check a sign change of f′ for a turning point.
1. For the can of volume 250π cm³, what radius minimises the surface area?
2. For y = x⁴, f″(0) = 0. The origin is:
3. For y = x³ − 6x² + 9x, what is the gradient at the point of inflexion?
In optimisation, forming the function in one variable using the constraint is usually two marks and most of the difficulty.
Justifying maximum or minimum. Either method is fine, but it has to be on the page.
For an inflexion, showing the sign of f″ actually changes. Solving f″ = 0 and stopping is an incomplete answer, and the x⁴ case is exactly why.
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