Calculus finds where a function levels off. It does not, by itself, find where the function is largest. On a restricted domain the answer can sit at an endpoint, where the derivative is nothing special — and most optimisation problems come with a restricted domain built in.
Commit to a prediction about where the largest value sits.
You have found every solution of $f'(x) = 0$ on the interval $[a, b]$ and classified each one. Without looking ahead say whether you now know the largest value $f$ takes on that interval, and if not, what is still missing.
Separate local from global, and find where global extrema can hide.
Three places a global extremum can be
We just saw how to find and classify every stationary point. That raises a question: "maximum" in an exam question usually means the largest value overall, so is a local maximum the same thing? This card answers it → no, and the difference is a comparison over the whole domain rather than over a neighbourhood.
A global minimum is defined the same way with the inequality reversed. Compare that with a local maximum, where $f(c) \geq f(t)$ only for $t$ near $c$. Every global extremum is local; the reverse fails constantly.
A local maximum is a claim about a neighbourhood. A global maximum is a claim about the entire domain. Calculus tests neighbourhoods, so it cannot on its own settle a global question.
Where a global extremum can sit. There are exactly three possibilities, and dp12 exists because two of them are routinely forgotten:
Global max: $f(c) \geq f(t)$ for EVERY $t$ in the domain. Local: only for $t$ near $c$.; A global extremum sits at a stationary point, an endpoint, OR a discontinuity. Only the first is found by solving $f'(x) = 0$.
Pause, copy the definition of a global maximum and the three places a global extremum can occur, into your book.
Because there are only three places to look, finding a global extremum is a short and completely reliable procedure: list every candidate, evaluate $f$ at all of them, and compare the numbers.
Worked through. Find the global maximum and minimum of $f(x) = x^3 - 3x$ on $[0, 3]$.
$f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$, and only $x = 1$ lies in $[0, 3]$. The endpoints are $0$ and $3$. There are no discontinuities.
Evaluating: $f(0) = 0$, $f(1) = 1 - 3 = -2$, $f(3) = 27 - 9 = 18$.
So the global minimum is $-2$ at $x = 1$ and the global maximum is $\mathbf{18}$ at the endpoint $x = 3$. A student who stopped after the stationary points would have missed the maximum by a factor of nine.
Candidate method: interior solutions of $f'(x) = 0$, plus endpoints, plus discontinuities. Evaluate $f$ at all of them and compare.; Comparing values makes classifying the stationary points unnecessary for a GLOBAL question.
Pause, copy the four-step candidate method for global extrema, and the note that comparing values makes local classification unnecessary, into your book.
Quick check: On $[0, 3]$, $f(x) = x^3 - 3x$ has $f(0) = 0$, $f(1) = -2$ and $f(3) = 18$. What is the global maximum?
Build the function, the constraint and the domain yourself.
We just saw how to optimise a function you are handed. That raises a question: an exam optimisation question hands you a situation, not a function, so how do you get from words to something differentiable? This card answers it → define variables, write the quantity to optimise, use the constraint to eliminate all but one variable, and state the domain.
The five-step method, and every one of the five is a marked step:
Step 5 matters too. If the question asks for the maximum area, an answer of "$x = 10$" is not the answer: it is the value of the variable that produces it. Read the question again before writing the final line.
Optimisation method: define variables, write the quantity, write the constraint, eliminate to ONE variable and state the domain, optimise and answer in context.; The domain comes from the physical situation, and it is what makes endpoint checks and rejections possible.
Pause, copy the five-step optimisation method, and the note that the domain comes from the physical situation and must be stated, into your book.
True or false: If you have found and classified every stationary point of $f$ on $[a, b]$, you have found the largest value $f$ takes there.
Work three optimisation problems in full.
Worked examples · 3 in a row, reveal as you go
A farmer has 40 m of fencing to enclose a rectangular pen, using an existing straight wall as one full side. Find the dimensions giving the greatest area, and state that area.
An open box is made from a 24 cm square of card by cutting equal squares of side $x$ cm from each corner and folding up the sides. Find $x$ giving the greatest volume.
Find the global maximum and minimum of $f(x) = x + \dfrac{4}{x}$ on the interval $\left[\tfrac{1}{2},\, 4\right]$.
Fill the blanks: 40 m of fencing, three sides of a rectangle against a wall, $x$ metres perpendicular to the wall.
$x$ at maximum area $=$
the other side $y =$
greatest area $=$
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · model, restrict, compare
Find the global maximum and minimum of $f(x) = x^2 - 4x + 1$ on $[0, 5]$.
A rectangle has perimeter 24 cm. Write its area as a function of one side, state the domain, and find the greatest area.
Explain why $f(x) = x^3$ has no global maximum on $\mathbb{R}$, but does have one on $[-2, 5]$.
$f(x) = |x - 3|$ on $[0, 10]$. Find the global minimum and explain why solving $f'(x) = 0$ never finds it.
A cylinder has a fixed volume. A student differentiates the surface area with respect to $r$ while $h$ is still in the expression. What is wrong, and what must be done first?
Odd one out: Three of these are places a global maximum can occur. Which one is not a place you need to check?
Match each step of an optimisation question to what it produces:
Earlier you were asked whether finding and classifying every solution of $f'(x) = 0$ on $[a, b]$ tells you the largest value. It does not. Two candidates are still missing: the endpoints $f(a)$ and $f(b)$, where the gradient is generally not zero so calculus never looks; and any point where $f$ is discontinuous or not differentiable, such as a corner. The reliable procedure is to list all three kinds of candidate, evaluate $f$ at each, and compare the numbers — at which point classifying the stationary points becomes unnecessary, because the comparison has already answered the question.
That completes the graphing and optimisation half of this focus area. The remaining lessons apply the same derivatives to quantities that change with time: interpreting first and second derivatives in context, recovering a quantity from its rate with an initial condition, proportional growth and decay, and the displacement–velocity–acceleration chain.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 1 for the checkpoint question bank.
Q1. Find the global maximum and global minimum of $f(x) = 2x^3 - 9x^2 + 12x$ on the interval $[0, 3]$.
(a) Find all stationary points in the interval. (b) List every candidate, including endpoints, and evaluate $f$ at each. (c) State the global maximum and minimum, with the $x$-value at which each occurs. (3 marks)
Q2. A rectangular field is to be enclosed and then divided into two equal rectangles by a fence parallel to one side. The total length of fencing available is 300 m.
(a) Define your variables, write the area to be maximised and the constraint, and state the domain. (b) Express the area as a function of one variable and find the dimensions that maximise it. (c) State the maximum area, and explain why the endpoints of your domain do not need separate evaluation. (3 marks)
Q3. A manufacturer models profit as $P(x)$ dollars for $x$ units produced per day, where $x$ must be a whole number between 0 and 500. Calculus gives a single stationary point at $x = 237.4$, which is a local maximum.
(a) Explain why $x = 237.4$ cannot be the answer, and describe exactly what you would do instead. (b) Explain why checking $x = 0$ and $x = 500$ is still necessary even though a local maximum exists. (c) A colleague argues that because $P$ is only defined at whole numbers, differentiating it is meaningless and the whole method is invalid. Evaluate that argument. (3 marks)
Drill 1: $f'(x) = 2x - 4 = 0$ gives $x = 2$, which is in $[0, 5]$. Candidates $0$, $2$, $5$: $f(0) = 1$, $f(2) = 4 - 8 + 1 = -3$, $f(5) = 25 - 20 + 1 = 6$. Global minimum $-3$ at $x = 2$; global maximum $6$ at the endpoint $x = 5$.
Drill 2: $2x + 2y = 24$ gives $y = 12 - x$, so $A(x) = x(12 - x) = 12x - x^2$ with domain $0 < x < 12$. $A'(x) = 12 - 2x = 0$ at $x = 6$, and $A''= -2 < 0$, a maximum. Then $y = 6$ and $A = 36$ cm². The rectangle of greatest area for a given perimeter is a square.
Drill 3: On $\mathbb{R}$, $x^3 \to \infty$ as $x \to \infty$, so no value is ever largest and there is no global maximum. On $[-2, 5]$ the candidates are the endpoints and the stationary point $x = 0$: $f(-2) = -8$, $f(0) = 0$, $f(5) = 125$. The global maximum is $125$ at the endpoint $x = 5$. Restricting the domain is exactly what makes a global maximum exist.
Drill 4: $|x - 3| \geq 0$ with equality only at $x = 3$, so the global minimum is $0$ at $x = 3$. Solving $f'(x) = 0$ never finds it because $f'$ does not exist at $x = 3$ — the graph has a corner, with gradient $-1$ on the left and $+1$ on the right. It is the third kind of candidate: a point where $f'$ fails to exist.
Drill 5: The problem is that $h$ is not independent of $r$: the fixed volume ties them together, so treating $h$ as a constant while differentiating with respect to $r$ is invalid. First use the constraint $V = \pi r^2 h$ to write $h = \dfrac{V}{\pi r^2}$, substitute that into the surface-area expression to get a function of $r$ alone, and only then differentiate.
Q1 (3 marks): (a) $f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$, so the stationary points are $x = 1$ and $x = 2$, and both lie in $[0, 3]$ [1]. (b) The candidates are $0$, $1$, $2$ and $3$. $f(0) = 0$; $f(1) = 2 - 9 + 12 = 5$; $f(2) = 16 - 36 + 24 = 4$; $f(3) = 54 - 81 + 36 = 9$ [1]. (c) The global maximum is $\mathbf{9}$ at $x = 3$, an endpoint, and the global minimum is $\mathbf{0}$ at $x = 0$, the other endpoint. Notice that both global extrema are at endpoints, and neither of the two stationary points is either of them — the local maximum at $x = 1$ gives only 5 [1].
Q2 (3 marks): (a) Let $x$ metres be the length of each of the three parallel fences (the two ends plus the divider) and $y$ metres the length of each of the two remaining sides. Area $A = xy$. The constraint is $3x + 2y = 300$. Domain: $x > 0$ and $y > 0$ forces $0 < x < 100$ [1]. (b) From the constraint $y = \dfrac{300 - 3x}{2}$, so $A(x) = x \cdot \dfrac{300 - 3x}{2} = 150x - \dfrac{3x^2}{2}$. Then $A'(x) = 150 - 3x = 0$ gives $x = 50$, and $A''(x) = -3 < 0$, confirming a maximum. So $y = \dfrac{300 - 150}{2} = 75$, and the field is $50$ m by $75$ m [1]. (c) The maximum area is $50 \times 75 = 3750$ m². The endpoints of the domain need no separate evaluation because the domain is open: as $x \to 0^+$ or $x \to 100^-$ one dimension vanishes and $A \to 0$, so neither end can supply a maximum. (Stating this is the check; it is not the same as skipping it.) [1]
Q3 (3 marks): (a) Production is a whole number of units, so $x = 237.4$ is not in the domain and cannot be the answer. Since $P$ is a smooth function with a single local maximum there, the best whole-number value is one of the two nearest integers, so evaluate $P(237)$ and $P(238)$ and take the larger. Rounding to $237$ without checking $238$ is not safe: the curve need not be symmetric about its peak [1]. (b) A local maximum only claims to beat its immediate neighbours. The global maximum over $0 \leq x \leq 500$ could sit at either end, so $P(0)$ and $P(500)$ must be compared with $P(237)$ and $P(238)$. In a profit model $P(0)$ is often negative (fixed costs) and $P(500)$ may be too, but that has to be shown rather than assumed [1]. (c) The colleague is right on the literal point: a function defined only on integers has no derivative, because the limit defining $f'$ requires values arbitrarily close to $x$. But the argument does not invalidate the method. What is being differentiated is a continuous model $P(x)$ fitted to the situation, and the model is a legitimate tool provided its conclusions are then translated back to the actual domain — which is precisely what part (a) does by testing the neighbouring integers. The right response is to be explicit that the continuous function is a model, not to abandon calculus; the alternative, evaluating all 501 integers, is exact but tells you nothing about why the optimum is where it is [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →