SAMath MethodsSyllabus dot point

How do we use calculus to find the maximum or minimum value of a real-world quantity?

Optimisation problems use the first derivative to locate the maximum or minimum value of a quantity subject to a constraint.

A step-by-step method for solving optimisation problems: build the function, use a constraint to reduce to one variable, differentiate, solve for stationary points, and justify the maximum or minimum.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The standard method
Reducing to one variable
Justifying the nature of the stationary point
A volume problem with a chain of variables
Common errors
Why it matters

What this dot point is asking

An optimisation problem asks for the largest or smallest possible value of some quantity - maximum volume, minimum cost, shortest distance, greatest area. Calculus solves these because a smooth maximum or minimum occurs where the gradient is zero, i.e. at a stationary point where $f'(x)=0$ .

The standard method

Almost every optimisation problem follows the same five steps:

Define variables and write the quantity to optimise as a formula. This is the objective function.
Use the constraint to express the objective in terms of a single variable.
Differentiate and set the derivative equal to zero.
Solve for the stationary point(s).
Justify that you have the required maximum or minimum (second-derivative test or sign of $f'$ ), and state the domain so you can also check endpoints.

Reducing to one variable

The step students find hardest is using the constraint. Suppose a rectangle has perimeter $40$ m, so $2x+2y=40$ , giving $y=20-x$ . The area $A=xy$ then becomes a function of $x$ alone:

A(x)=x(20-x)=20x-x^2.

Now there is only one variable to differentiate with respect to.

Maximum area of a rectangular paddock

Worked example

A farmer has $40$ m of fencing to enclose a rectangular paddock. What dimensions give the maximum area?

Let the sides be $x$ and $y$ . The constraint is the perimeter:

2x+2y=40 \quad\Rightarrow\quad y=20-x.

The objective is the area:

A(x)=x(20-x)=20x-x^2, \qquad 0<x<20.

Differentiate and set to zero:

A'(x)=20-2x=0 \quad\Rightarrow\quad x=10.

Confirm it is a maximum with the second derivative:

A''(x)=-2<0,

x=10

gives a maximum. Then

y=20-10=10

The maximum area is a $10\text{ m}\times 10\text{ m}$ square with area $A=100\ \text{m}^2$ .

Justifying the nature of the stationary point

You must show why your answer is a maximum (or minimum) - marks are lost for finding $x$ but not justifying it.

Second-derivative test: if $f''(x)<0$ at the stationary point it is a local maximum; if $f''(x)>0$ it is a local minimum.
Sign test: check the sign of $f'(x)$ just before and just after the stationary point. A change from $+$ to $-$ is a maximum; $-$ to $+$ is a minimum.

A volume problem with a chain of variables

Largest open box from a square sheet

Worked example

A square sheet of card of side $12$ cm has equal squares of side $x$ cut from each corner; the sides are folded up to make an open box. Find $x$ that maximises the volume.

The base is $(12-2x)$ by $(12-2x)$ and the height is $x$ , so

V(x)=x(12-2x)^2, \qquad 0<x<6.

Expand: $V=x(144-48x+4x^2)=144x-48x^2+4x^3$ . Differentiate:

V'(x)=144-96x+12x^2=12(x^2-8x+12)=12(x-2)(x-6).

Setting $V'(x)=0$ gives $x=2$ or $x=6$ . Since $0<x<6$ , only $x=2$ is valid. Check:

V''(x)=-96+24x, \qquad V''(2)=-96+48=-48<0,

x=2

is a maximum. The maximum volume is

V(2)=2(12-4)^2=2\times 64=128\ \text{cm}^3.

Common errors

Why it matters

Optimisation is the headline application of differentiation in Stage 2 and a reliable source of extended-response marks in both the Skills and Applications Tasks and the external exam. The same five-step structure - model, constrain, differentiate, solve, justify - recurs in your Mathematical Investigation, where you may optimise a quantity in a modelling context of your own choosing.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2018 SACE Stage 23 marksFor f(x) = x e^(-x) (x >= 0), the tangent to the graph at the point where x = a has y-intercept Y(a) = a^2 e^(-a). Determine the value of a that maximises the y-intercept of the tangent.

Show worked answer →

We optimise Y(a) = a^2 e^(-a) over a >= 0 using calculus.

Differentiate with the product rule (treating a as the variable). Let u = a^2 and v = e^(-a), so u' = 2a and v' = -e^(-a):
Y'(a) = 2a e^(-a) + a^2 (-e^(-a)) = e^(-a) (2a - a^2) = a e^(-a) (2 - a).
Set Y'(a) = 0. Since e^(-a) > 0 always, the stationary points are a = 0 and a = 2.
Justify which is the maximum. For 0 < a < 2, (2 - a) > 0 so Y'(a) > 0 (increasing); for a > 2, (2 - a) < 0 so Y'(a) < 0 (decreasing). The gradient changes from positive to negative at a = 2, confirming a local maximum.

So the y-intercept of the tangent is maximised at a = 2. Marks: one for differentiating correctly, one for solving Y'(a) = 0, and one for justifying that a = 2 gives the maximum.