Maxima & Minima Notes

Last reviewed 16 Sept 2026 · 7 min read

Increasing and decreasing functions

$f^{'} (x) > 0$ on an interval → $f$ is increasing; $f^{'} (x) < 0$ → decreasing.
Critical (stationary) points — where $f^{'} (x) = 0$ (or $f^{'}$ does not exist).

Local and absolute extrema

Local (relative) maximum at $c$ : $f (c) \geq f (x)$ for $x$ near $c$ ; local minimum similarly.
Absolute (global) maximum/minimum — largest/smallest value over the whole domain.

First derivative test

At a critical point $c$ :

$f^{'}$ changes from + to − → local maximum.
$f^{'}$ changes from − to + → local minimum.
No sign change → neither (possible inflection).

Second derivative test

At $c$ with $f^{'} (c) = 0$ :

$f^{''} (c) < 0$ → local maximum.
$f^{''} (c) > 0$ → local minimum.
$f^{''} (c) = 0$ → test fails — use the first derivative test or higher derivatives (if the first non-zero derivative at $c$ is of even order, extremum: max if negative, min if positive; if odd order, inflection point).

Concavity and inflection

$f^{''} > 0$ → concave upward (convex); $f^{''} < 0$ → concave downward.
Point of inflection — where concavity changes ( $f^{''} = 0$ and changes sign).

Absolute extrema on a closed interval [a, b]

Evaluate $f$ at all critical points in (a, b) and at the end points a and b; the largest is the absolute maximum, the smallest the absolute minimum.

Functions of two variables

For $z = f (x, y)$ :

Stationary points: solve $f_{x} = 0$ and $f_{y} = 0$ .
At each point compute $r = f_{xx}$ , $s = f_{x y}$ , $t = f_{y y}$ .

FormulaSecond derivative test (two variables)

Condition	Nature
$r t - s^{2} > 0$ and $r < 0$	Maximum
$r t - s^{2} > 0$ and $r > 0$	Minimum
$r t - s^{2} < 0$	Saddle point (neither)
$r t - s^{2} = 0$	Inconclusive — further investigation

Lagrange multipliers (constrained optimisation)

To find extrema of $f (x, y, z)$ subject to $ϕ (x, y, z) = 0$ :

Form $F = f + λ ϕ$ .
Solve $F_{x} = 0$ , $F_{y} = 0$ , $F_{z} = 0$ and $ϕ = 0$ for $x, y, z, λ$ .
Evaluate $f$ at the solutions to identify maxima/minima (by physical reasoning or further tests).

(Equivalent condition: $\nabla f = - λ \nabla ϕ$ — gradients are parallel at the optimum.)

Engineering applications

Problem	Result
Rectangle of given perimeter with maximum area	Square
Rectangle of given area with minimum perimeter	Square
Open box from a square sheet of side $a$ by cutting squares of side $x$ from corners	Maximum volume at $x = a /6$ , $V_{ma x} = 2 a^{3} /27$
Cylinder of given volume with minimum total surface area (closed)	Height = diameter
Strongest rectangular beam (maximum section modulus $b d^{2} /6$ ) cut from a circular log of diameter $D$	$b = D / 3$ , $d = D 2/3$ → $d / b = 2$
Stiffest rectangular beam (maximum $b d^{3}$ ) from a log	$d / b = 3$
Most economical rectangular channel (max flow for given area)	Width = 2 × depth (see Open Channel Flow)
Bending moment maximum	Where shear force = 0 ( $d M / d x = V = 0$ )

Worked examples

Worked ExampleExample 1 — local extrema

Find the local maxima and minima of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

Solution. $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ → critical points $x = 1, 3$ $f^{''} (x) = 6 x - 12$ : $f^{''} (1) = - 6 < 0$ → maximum $f (1) = 5$ ; $f^{''} (3) = 6 > 0$ → minimum $f (3) = 1$ Inflection at $x = 2$ ( $f^{''} = 0$ , sign change).

Worked ExampleExample 2 — absolute extrema on an interval

Find the absolute extrema of the same function on $[0, 4]$ .

Solution. $f (0) = 1$ , $f (1) = 5$ , $f (3) = 1$ , $f (4) = 64 - 96 + 36 + 1 = 5$ Absolute maximum 5 (at $x = 1$ and $x = 4$ ); absolute minimum 1 (at $x = 0$ and $x = 3$ )

Worked ExampleExample 3 — two variables

Find and classify the stationary points of $f (x, y) = x^{3} + y^{3} - 3 x y$ .

Solution. $f_{x} = 3 x^{2} - 3 y = 0$ , $f_{y} = 3 y^{2} - 3 x = 0$ → $y = x^{2}$ , $x = y^{2}$ → points (0, 0) and (1, 1) $r = 6 x$ , $s = - 3$ , $t = 6 y$ At (0, 0): $r t - s^{2} = - 9 < 0$ → saddle point At (1, 1): $r t - s^{2} = 36 - 9 = 27 > 0$ , $r = 6 > 0$ → minimum, $f = - 1$

Worked ExampleExample 4 — simple minimum

Find the minimum of $f (x, y) = x^{2} + y^{2} - 4 x + 6 y + 5$ .

Solution. $f_{x} = 2 x - 4 = 0$ , $f_{y} = 2 y + 6 = 0$ → (2, −3); $r = 2$ , $t = 2$ , $s = 0$ → $r t - s^{2} = 4 > 0$ , $r > 0$ → minimum $f = 4 + 9 - 8 - 18 + 5 =$ −8

Worked ExampleExample 5 — open box

Squares are cut from the corners of a 12 cm × 12 cm sheet to make an open box. Find the maximum volume.

Solution. $V = x (12 - 2 x)^{2}$ → $V^{'} = (12 - 2 x) (12 - 6 x) = 0$ → $x = 2$ (x = 6 gives zero volume) $V_{ma x} = 2 \times 8^{2} =$ 128 cm³ (check $2 a^{3} /27 = 2 \times 1728/27 = 128$ ✓)

Worked ExampleExample 6 — Lagrange multipliers

Maximise $x y$ subject to $x + y = 10$ .

Solution. $F = x y + λ (x + y - 10)$ : $F_{x} = y + λ = 0$ , $F_{y} = x + λ = 0$ → $x = y = 5$ → maximum 25

Worked ExampleExample 7 — strongest beam from a log

A rectangular beam is cut from a log of 300 mm diameter. Find $b$ and $d$ for maximum bending strength.

Solution. Maximise $S = b d^{2} = b (D^{2} - b^{2})$ : $d S / d b = D^{2} - 3 b^{2} = 0$ → $b = D / 3 =$ 173.2 mm, $d = D^{2} - b^{2} = 90000 - 30000 =$ 244.9 mm ( $d / b = 2$ )

Worked ExampleExample 8 — minimum surface cylinder

A closed cylindrical tank must hold 1000 m³. Find the dimensions for minimum surface area.

Solution. $A = 2 π r^{2} + 2 π r h$ with $π r^{2} h = 1000$ → $A = 2 π r^{2} + \frac{2000}{r}$ → $A^{'} = 4 π r - \frac{2000}{r ^{2}} = 0$ → $r^{3} = \frac{500}{π}$ → $r =$ 5.42 m; $h = \frac{1000}{π r ^{2}} =$ 10.84 m ( $h = 2 r$ )

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Maxima & Minima