Maxima and minima of single-variable functions

At a Glance

• Frequency: 7 sub-parts across 6 of 13 years (2014, 2018, 2019, 2020, 2024, 2025)
• Priority tier: T2
• Marks (count): 10 (1), 13 (2), 15 (1), 20 (3)
• Average solve time: ~12 min
• Difficulty mix: medium 4, hard 2, easy 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Single-variable optimisation appears in 6 of the last 13 years at 10–20 marks, almost entirely in Section A. The three archetypes rotate: applied optimisation (geometry and word problems reduced to one variable), critical-point analysis on FTC-defined integrals, and extrema on closed intervals. The applied-optimisation type dominates (4 of 7 questions, including the 20-mark 2024 question) and requires a clear geometric setup before any calculus. The FTC-based critical-point analysis (2020, 20 marks) is the hardest variant — it demands sign-table reasoning across four roots. The closed-interval extremum is the most mechanical: evaluate at critical points and both endpoints.

Minimum Theory

Critical points and the first-derivative test. $f'(c)=0$ (or $f'(c)$ does not exist) at a critical point $c$. If $f'$ changes from positive to negative at $c$, then $c$ is a local maximum; negative to positive gives a local minimum.

Second-derivative test. If $f'(c)=0$ and $f''(c)<0$, then $c$ is a local maximum; $f''(c)>0$ gives a local minimum; $f''(c)=0$ is inconclusive.

Closed-interval extremum (Extreme Value Theorem). On $[a,b]$, the absolute maximum and minimum are attained at either a critical point interior to $(a,b)$ or at an endpoint. Evaluate $f$ at all critical points in $(a,b)$ and at $a,b$; the largest value is the maximum, the smallest is the minimum.

FTC-defined functions. If $f(x)=\int_a^x g(t)\,dt$, then by the FTC, $f'(x)=g(x)$. To find critical points of $f$, solve $g(x)=0$; classify by the sign of $g$ (which is $f'$) on either side.

Applied optimisation — standard setup. Identify the quantity to optimise and the geometric/physical constraint; use the constraint to reduce to one variable; differentiate and solve; check endpoints and that the critical point is the correct type.

Question Archetypes

Three patterns cover every single-variable optimisation question in the corpus.

applied-optimization (4 question(s); 2014, 2018, 2024, 2025)

Recognition Cues

• “Find the [greatest/maximum/minimum] [volume/distance/height/side] of [a cylinder/box/distance from a point/shape].”
• A geometric constraint (inscribed in a sphere, inscribed in a cone, folded from a sheet) links two variables; one substitution leaves a one-variable problem.
• The 2024 question (20 marks) is “find the volume of the greatest cylinder inscribed in a cone” — the standard hard variant.

Solution Template

1. Draw and label. Assign one free variable (e.g. radius $r$ or height $h$).
2. Write the constraint. Use similar triangles, Pythagoras, or the geometric relationship to express the other dimension in terms of the free variable.
3. Express the objective function $V(r)$ or $D(r)$ in one variable.
4. Differentiate and set to zero. Solve for the critical point(s).
5. Confirm maximum via $V''<0$ or endpoint values.
6. State the answer (value of the optimised quantity and, if asked, the corresponding dimensions).

Worked Example(s)

2014 Paper 1, 2014-P1-Q3a (15 marks)

Find the height of the cylinder of maximum volume inscribed in a sphere of radius $a$.

Let half-height be $h$; base radius $r$ satisfies $r^2+h^2=a^2$. Volume $V=\pi r^2\cdot2h=2\pi h(a^2-h^2)$.

$V'=2\pi(a^2-3h^2)=0\Rightarrow h=a/\sqrt3$. $V''=-12\pi h<0$: maximum.

$\boxed{\text{Height}=2h=\frac{2a}{\sqrt3}=\frac{2a\sqrt3}{3}.}$

2024 Paper 1, 2024-P1-Q3b (20 marks)

Find the volume of the greatest cylinder inscribed in a cone of height $h$ and semi-vertical angle $\alpha$.

Cylinder radius $r$, base at distance $h-H$ from apex (where $H=$ cylinder height). Similar triangles: $r=H\tan\alpha$… Let the cylinder top be at distance $x$ from the apex: $r=x\tan\alpha$, height $=h-x$.

$V=\pi r^2(h-x)=\pi(x\tan\alpha)^2(h-x)=\pi\tan^2\alpha\,(hx^2-x^3)$.

$dV/dx=\pi\tan^2\alpha(2hx-3x^2)=0$ at $x=2h/3$.

$H^*=h-2h/3=h/3$. $r^*=\tfrac{2h}{3}\tan\alpha$.

$V^*=\pi\tan^2\alpha\cdot\frac{4h^2}{9}\cdot\frac{h}{3}=\boxed{\frac{4\pi h^3\tan^2\alpha}{27}.}$

Note: cylinder height is always $h/3$, independent of $\alpha$.

2018 Paper 1, 2018-P1-Q2b (13 marks)

Find the shortest distance from $(1,0)$ to the parabola $y^2=4x$.

Parametrise: points on $y^2=4x$ are $(t^2,2t)$. Squared distance: $D(t)=(t^2-1)^2+4t^2=(t^2+1)^2$. Minimised at $t=0$ giving $D_{\min}=1$.

$\boxed{\text{Shortest distance}=1,\text{ attained at the vertex }(0,0).}$

The perfect-square collapse $(t^2-1)^2+4t^2=(t^2+1)^2$ is the key insight; $(1,0)$ is the focus.

2025 Paper 1, 2025-P1-Q1c (10 marks)

Rectangular sheet $6\times2$ m; equal squares cut from corners; fold to open box. Find height for maximum volume.

Height $=x$; base $=(6-2x)\times(2-2x)$. $V=4x^3-16x^2+12x$, $0<x<1$.

$V'=12x^2-32x+12=0\Rightarrow x=(4\pm\sqrt7)/3$. Only $x=(4-\sqrt7)/3\approx0.45$ m lies in $(0,1)$. $V''<0$: maximum.

Common Traps

• For cylinder-in-sphere: confuse half-height $h$ with full height $2h$; the constraint is $r^2+h^2=a^2$ (not $r^2+(2h)^2=a^2$).
• For cylinder-in-cone: the top circle (not the base) touches the cone wall. Similar triangles give $r/x=\tan\alpha$ where $x$ is measured from the apex, not from the base.
• For the parabola problem: the squared distance collapses to a perfect square $(t^2+1)^2$ — if you expand naively instead of recognising the pattern, you get unnecessary algebra.

critical-point-analysis (1 question(s); 2020)

Recognition Cues

• $f(x)=\int_0^x g(t)\,dt$ where $g$ is a product of quadratics; “find critical points, local maxima, local minima, zeros on $[0,5]$.”
• FTC immediately gives $f'(x)=g(x)$; the problem reduces to analysing the sign of $g$.

Solution Template

1. $f'=g$ by FTC. Write out $g(x)$ and factor into linear (or simple quadratic) terms.
2. Sign table of $f'$. Identify all roots; determine the sign of $f'$ in each interval between roots.
3. Classify. Local max at each $+\to-$ sign change; local min at each $-\to+$ change.
4. Zeros of $f$. Compute $f$ at all critical points and endpoints; check whether $f$ returns to zero.

Worked Example(s)

2020 Paper 1, 2020-P1-Q3a (20 marks)

$f(x)=\int_0^x(t^2-5t+4)(t^2-5t+6)\,dt$. (i) critical points; (ii) local minima; (iii) local maxima; (iv) number of zeros on $[0,5]$.

$f'(x)=(x-1)(x-2)(x-3)(x-4)$ (factor each quadratic).

Sign table (alternating from + for $x<1$): $+,-,+,-,+$.

• Local max at $x=1,3$ ($+\to-$); local min at $x=2,4$ ($-\to+$).

$f(0)=0$; $f(1)\approx7.87$ (local max); $f(2)\approx7.73$ (local min); $f(3)\approx8.10$ (local max); $f(4)\approx7.47$ (local min); $f(5)\approx15.83$.

Both local minima are $>0$, so $f>0$ on $(0,5]$.

$\boxed{\text{Exactly 1 zero on }[0,5]\text{ (at }x=0\text{).}}$

Common Traps

• With four roots the sign pattern is $+,-,+,-,+$ starting from the left — it alternates at each root. This means $x=1$ and $x=3$ are maxima; $x=2$ and $x=4$ are minima.
• The zero count requires evaluating $f$ at the interior minima (not just checking the formula): all minima exceed $0$, so $f$ never dips back to zero after $x=0$.
• $f(0)=0$ by the FTC (lower limit is $0$), not a critical point of $f$ (since $f'(0)=24>0$); it is the only zero.

extrema-on-interval (2 question(s); 2018, 2019)

Recognition Cues

• “Find the maximum and minimum of $f(x)$ on $[a,b]$.” The interval is explicitly closed and bounded.
• The critical point(s) of $f$ may or may not lie inside $[a,b]$; endpoints always must be evaluated.

Solution Template

1. Find $f'(x)$ and solve $f'(x)=0$.
2. Discard critical points outside $[a,b]$.
3. Evaluate $f$ at all remaining critical points and at $a$ and $b$.
4. Report the largest value as the maximum and the smallest as the minimum, with their locations.

Worked Example(s)

2019 Paper 1, 2019-P1-Q3a (15 marks)

Find the maximum and minimum of $f(x)=2x^3-9x^2+12x+6$ on $[2,3]$.

$f'=6(x-1)(x-2)=0$ at $x=1,2$. On $[2,3]$: only $x=2$ (left endpoint) qualifies.

For $x\in(2,3)$: $f'>0$ (both factors positive), so $f$ is strictly increasing.

$f(2)=16-36+24+6=10$; $f(3)=54-81+36+6=15$.

$\boxed{\text{Min}=10\text{ at }x=2;\quad\text{Max}=15\text{ at }x=3.}$

2018 Paper 1, 2018-P1-Q4a (13 marks)

Find max and min of $f(x)=x^4-5x^2+4$ on $[2,3]$.

$f'=4x^3-10x=2x(2x^2-5)=0$ at $x=0,\pm\sqrt{5/2}\approx\pm1.58$. None lie in $(2,3)$.

For $x\in[2,3]$: $f'(2)=32-20=12>0$, so $f$ is increasing on $[2,3]$.

$f(2)=16-20+4=0$; $f(3)=81-45+4=40$.

$\boxed{\text{Min}=0\text{ at }x=2;\quad\text{Max}=40\text{ at }x=3.}$

Common Traps

• Critical points outside $[a,b]$ are irrelevant — always check that each critical point lies in the interior $(a,b)$ before evaluating.
• On a closed interval where $f'$ doesn’t vanish (e.g. $f$ is monotone throughout), the extrema are both at endpoints; no critical-point calculation is needed beyond confirming monotonicity.
• Report the value of $f$ at the extremum, not just the location. The question asks for “the maximum value” — giving only $x=3$ without $f(3)=15$ loses marks.

Practice Set

• 2014-P1-Q3a (15 m) — — cylinder-in-sphere: express $V(h)$ using the sphere constraint; standard template.
• 2020-P1-Q3a (20 m) — — FTC-defined $f$, four critical points, sign table, zero count — the hardest type in this chapter.