A stationary point of \(y=\mathrm{f}(x)\) is a point where \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\). Solving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\) only finds the candidates — a separate test decides whether each one is a maximum point, a minimum point or a stationary point of inflexion. Maximum point Minimum point Stationary point of inflexion Second-derivative test at a stationary point \(x=a\) Value of \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x=a\)Conclusion \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}<0\)Maximum point \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}>0\)Minimum point \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0\)Inconclusive — use the first-derivative test First-derivative test: sign of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as \(x\) increases through \(a\) Nature\(x=a^{-}\)\(x=a\)\(x=a^{+}\) Maximum point\(+\) \(\nearrow\)\(0\)\(-\) \(\searrow\) Minimum point\(-\) \(\searrow\)\(0\)\(+\) \(\nearrow\) Possible stationary inflexion; check concavity\(+\) \(\nearrow\)\(0\)\(+\) \(\nearrow\) A zero second derivative makes the second-derivative test inconclusive. A first derivative with no sign change rules out a local maximum or minimum under the displayed sign conditions, but does not alone prove inflexion. To establish a stationary point of inflexion, also verify a change in concavity, for example a sign change of the second derivative on the adjacent intervals. Maxima and minima problems Method Define the variables and write down the quantity to be maximised or minimised. Use the given constraint so that the quantity contains one variable only. Differentiate and solve the stationary condition. Justify the nature with the second-derivative test (or a sign test). Answer the question asked, with units. Mensuration formulas that often appear SolidVolumeCurved surface area Cylinder\(\pi r^2h\)\(2\pi rh\) Cone\(\dfrac{1}{3}\pi r^2h\)\(\pi rl\) Sphere\(\dfrac{4}{3}\pi r^3\)\(4\pi r^2\) Exam check Keywords: greatest, largest, fastest \(\Rightarrow\) maximum; least, smallest, slowest \(\Rightarrow\) minimum. Before differentiating, express the quantity to optimise in one independent variable, using the constraint to eliminate the other variables. A zero second derivative makes the second-derivative test inconclusive. A first derivative with no sign change rules out a local maximum or minimum under the displayed sign conditions, but does not alone prove inflexion. To establish a stationary point of inflexion, also verify a change in concavity, for example a sign change of the second derivative on the adjacent intervals. Before differentiating, state the feasible domain from the geometry or context. Check any admissible endpoints and relevant limiting values as well as interior stationary points. A stationary value outside the feasible domain is not a physical optimum; an open endpoint may give an infimum rather than an attained minimum.

Stationary Points, Maxima and Minima

Secondary 4

TGM Original Questions

A stationary point of \(y=\mathrm{f}(x)\) is a point where \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\). Solving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\) only finds the candidates — a separate test decides whether each one is a maximum point, a minimum point or a stationary point of inflexion.

Maximum point

Minimum point

Stationary point of inflexion

Second-derivative test at a stationary point \(x=a\)

Value of \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x=a\)	Conclusion
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}<0\)	Maximum point
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}>0\)	Minimum point
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0\)	Inconclusive — use the first-derivative test

First-derivative test: sign of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as \(x\) increases through \(a\)

Nature	\(x=a^{-}\)	\(x=a\)	\(x=a^{+}\)
Maximum point	\(+\) \(\nearrow\)	\(0\)	\(-\) \(\searrow\)
Minimum point	\(-\) \(\searrow\)	\(0\)	\(+\) \(\nearrow\)
Possible stationary inflexion; check concavity	\(+\) \(\nearrow\)	\(0\)	\(+\) \(\nearrow\)

A zero second derivative makes the second-derivative test inconclusive. A first derivative with no sign change rules out a local maximum or minimum under the displayed sign conditions, but does not alone prove inflexion. To establish a stationary point of inflexion, also verify a change in concavity, for example a sign change of the second derivative on the adjacent intervals.

Maxima and minima problems

Method

Define the variables and write down the quantity to be maximised or minimised.
Use the given constraint so that the quantity contains one variable only.
Differentiate and solve the stationary condition.
Justify the nature with the second-derivative test (or a sign test).
Answer the question asked, with units.

Mensuration formulas that often appear

Solid	Volume	Curved surface area
Cylinder	\(\pi r^2h\)	\(2\pi rh\)
Cone	\(\dfrac{1}{3}\pi r^2h\)	\(\pi rl\)
Sphere	\(\dfrac{4}{3}\pi r^3\)	\(4\pi r^2\)

Exam check

Keywords: greatest, largest, fastest \(\Rightarrow\) maximum; least, smallest, slowest \(\Rightarrow\) minimum. Before differentiating, express the quantity to optimise in one independent variable, using the constraint to eliminate the other variables.

Before differentiating, state the feasible domain from the geometry or context. Check any admissible endpoints and relevant limiting values as well as interior stationary points. A stationary value outside the feasible domain is not a physical optimum; an open endpoint may give an infimum rather than an attained minimum.

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