Maxima and Minima

MATMaxima & Minima

This subtopic uses derivatives to find the greatest and least values of a function, both locally (turning points) and globally (absolute extrema on a domain). The core method is to locate critical points where $f'(x)=0$ (or $f'$ is undefined), then classify them using the first- or second-derivative test.

It is the highest-weight section of the chapter because of optimisation word-problems (boxes, cylinders, cones, windows, tanks) where you must build the objective function, reduce it to one variable using a constraint, and maximise or minimise it.

Stationary (critical) point

f'(x)=0

x=c

is a critical point of

f

f'(c)=0

f'(c)

does not exist; extrema occur only among such points.

Second derivative test

f'(c)=0,\ f''(c)<0 \Rightarrow \text{local max};\quad f''(c)>0 \Rightarrow \text{local min}

c

is a critical point. If

f''(c)=0

the test fails and you must use the first-derivative (sign-change) test.

First derivative (sign change) test

f'\ \text{changes} + \to - \text{ at } c \Rightarrow \text{max};\quad - \to + \Rightarrow \text{min}

Examine the sign of

f'(x)

just to the left and right of

c

; no sign change means

c

is neither (point of inflection).

Absolute extrema on a closed interval

M=\max\{f(c_i),f(a),f(b)\},\quad m=\min\{f(c_i),f(a),f(b)\}

[a,b]

, compare

f

at all critical points

c_i

inside

(a,b)

and at the endpoints

a,b

Method for word problems: (1) draw a figure and name variables, (2) write the quantity to optimise, (3) use the given constraint to reduce it to ONE variable, (4) set the derivative to $0$ , (5) confirm max/min by the second-derivative test, (6) state the answer with units.
Sphere–cylinder result: the cylinder of greatest volume inscribed in a sphere of radius $R$ has height $\dfrac{2R}{\sqrt{3}}$ .
Cone of given slant height has maximum volume when its semi-vertical angle is $\tan^{-1}\sqrt{2}$ .
Cylinder inscribed in a cone (height $H$ , base radius $R$ ) has greatest volume at radius $\dfrac{2R}{3}$ , and that volume is $\dfrac{4}{9}$ of the cone's volume.
Local maximum value is not always greater than a local minimum value of the same function; 'local' is relative to a neighbourhood only.
If $f''(c)=0$ , the second-derivative test is inconclusive — switch to the first-derivative test rather than guessing.
Discard roots of $f'(x)=0$ that fall outside the physical domain (e.g. a negative length or a cut larger than half the sheet).
For a sum constraint $x+y=k$ , optimise by substituting $y=k-x$ so the objective depends on a single variable.

Where the marks go

Stopping at $f'(x)=0$ without classifying the point — every critical point must be tested for max, min, or neither.
Forgetting endpoint values: on a closed interval the absolute maximum/minimum may occur at $a$ or $b$ , not at an interior critical point.
Maximising a quantity that should be minimised (or vice-versa) — read whether the problem wants greatest volume or least surface area.
Keeping geometrically impossible roots, e.g. a corner cut $x>9$ for an $18$ cm sheet, instead of rejecting them.

How the board asks it

Applicationthe optimisation method for word problems
A closed right circular cylinder of given total surface area $S$ is to hold the maximum volume. Show that its height is equal to the diameter of its base.
Numericalabsolute extrema on a closed interval
Find the absolute maximum and absolute minimum values of $f(x)=2x^3-15x^2+36x+1$ on the interval $[1,5]$ .
Numericalfirst- and second-derivative test classification
Find all the points of local maxima and local minima of the function $f(x)=x^4-62x^2+120x+9$ and the corresponding local extreme values.
Applicationdiscarding roots outside the physical domain
A square sheet of tin of side $18$ cm has a square of side $x$ cut from each corner; the flaps are folded up to form an open box. Find $x$ so that the volume of the box is maximum, and find that maximum volume.
Derive / provecone inscribed in a sphere result
Prove that the right circular cone of maximum volume that can be inscribed in a sphere of radius $R$ has altitude $\dfrac{4R}{3}$ .
Applicationthe sum-constraint substitution method
Divide the number $24$ into two positive parts such that the product of one part and the square of the other is a maximum.

Practise this topic

PreviousIncreasing and Decreasing Functions NextRolle’s Theorem and Lagrange’s Mean Value Theorem

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.