What is the largest area you can fence with a fixed length of wire? What price should a company charge to maximise profit? These are optimisation problems — and calculus gives us a systematic way to solve them. In this lesson, you will turn real-world scenarios into mathematical models, then use derivatives to find the best possible outcome.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
You have 20 metres of fencing and want to enclose a rectangular vegetable garden. One side of the garden will be against your house, so it does not need fencing. What dimensions do you think would give you the largest possible area? Make an initial guess and explain your reasoning.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
Optimisation problems ask us to find the maximum or minimum value of some quantity subject to certain restrictions. Here is a reliable six-step method:
In many real-world problems, the variable is restricted to a closed interval — for example, a length cannot be negative. The maximum or minimum might occur at a stationary point or at an endpoint of the domain.
When the domain is closed, check:
The largest of these values is the maximum; the smallest is the minimum.
🧮 Worked Examples
🧪 Activities
1 A rectangular garden is to be fenced on three sides with 24 metres of fencing (the fourth side is a wall). Find the dimensions that maximise the area.
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2 The sum of two positive numbers is 20. Find the numbers if the sum of their squares is to be as small as possible.
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3 Find the maximum and minimum values of $f(x) = x^3 - 3x + 1$ on $[0, 2]$.
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4 A company's profit function is $P(x) = -x^2 + 40x - 300$, where $x$ is the number of items sold. How many items should be sold to maximise profit?
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1 In an area maximisation problem, why must we express the area as a function of one variable before differentiating?
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2 A student finds a stationary point that gives a negative length. What does this tell them about their model, and what should they do next?
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Earlier you guessed the dimensions for a 20-metre fence against a wall.
Let the width be $x$ and the length be $y = 20 - 2x$. The area is $A = x(20 - 2x) = 20x - 2x^2$. Differentiating: $\frac{dA}{dx} = 20 - 4x = 0$, so $x = 5$ and $y = 10$. The maximum area is $50$ m$^2$. The optimal shape is not a square (because one side is free), but rather the length is twice the width. This is the power of calculus — it finds the exact answer without guessing.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. An open-top box is to be made from a rectangular sheet of cardboard measuring 30 cm by 20 cm by cutting identical squares of side length $x$ from each corner and folding up the sides. 4 MARKS
(a) Show that the volume of the box is $V = x(30 - 2x)(20 - 2x)$.
(b) Find the value of $x$ that maximises the volume, and state the maximum volume.
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Answer in your workbook.
7. A manufacturer finds that the cost of making $x$ units of a product is $C(x) = 0.5x^2 + 20x + 400$ dollars, and the revenue from selling $x$ units is $R(x) = 80x$ dollars. 4 MARKS
(a) Find the profit function $P(x)$.
(b) Find the production level that maximises profit, and calculate the maximum profit.
(c) If the factory can produce at most 50 units, find the maximum possible profit.
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Answer in your workbook.
8. Find the maximum and minimum values of $f(x) = 2x^3 - 3x^2 - 12x + 5$ on the interval $[-2, 3]$. Show all working, including endpoint checks. 4 MARKS
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1. $A = x(24 - 2x) = 24x - 2x^2$. $\frac{dA}{dx} = 24 - 4x = 0 \Rightarrow x = 6$. Dimensions: $6$ m by $12$ m. Max area $= 72$ m$^2$.
2. Let numbers be $x$ and $20 - x$. $S = x^2 + (20 - x)^2 = 2x^2 - 40x + 400$. $\frac{dS}{dx} = 4x - 40 = 0 \Rightarrow x = 10$. Both numbers are $10$.
3. $f'(x) = 3x^2 - 3 = 0 \Rightarrow x = 1$ (in $[0,2]$). $f(0) = 1$, $f(1) = -1$, $f(2) = 3$. Maximum $= 3$, minimum $= -1$.
4. $P'(x) = -2x + 40 = 0 \Rightarrow x = 20$. $P''(x) = -2 < 0$, so $20$ items maximises profit.
1. Differentiation finds stationary points of a function of one variable. If two variables remain, we do not have a standard derivative that can be set to zero.
2. A negative length is not physically possible. The student should check their domain restriction and possibly test the valid endpoint instead.
1. B — Before differentiating, write the objective function using one independent variable by using any given constraints.
2. A — Let width $= w$, then length $= 20 - w$. Area $A = w(20 - w)$, which is maximised at $w = 10$, giving $A = 100$ m$^2$.
3. A — $P'(x) = -4x + 100 = 0 \Rightarrow x = 25$.
4. A — Critical points at $x = 0$ and $x = 2$ (endpoints) and $x = 0$ from $f'(x)=0$. Evaluating: $f(0)=0$, $f(2)=-4$. The maximum is $0$.
5. B — Profit = Revenue - Cost.
6. B — Extrema on a closed interval occur at critical points (where $f'(x)=0$ or undefined) or at the endpoints.
7. B — Let base side $= x$ and height $= h$. Surface area: $x^2 + 4xh = 12$, so $h = \frac{12 - x^2}{4x}$. Volume $V = x^2 h = \frac{x(12-x^2)}{4}$. Maximising gives $x = 2$ and $V = 4$ m$^3$.
Q6 (4 marks): Cutting squares of side $x$ gives a base of $(30-2x)$ by $(20-2x)$ with height $x$, so $V = x(30-2x)(20-2x)$ [1]. Expanding: $V = 4x^3 - 100x^2 + 600x$, so $\frac{dV}{dx} = 12x^2 - 200x + 600$ [1]. Setting $\frac{dV}{dx} = 0$: $3x^2 - 50x + 150 = 0 \Rightarrow x = \frac{25 \pm 5\sqrt{7}}{3}$ [1]. Reject the larger root (as $x < 10$), so $x = \frac{25 - 5\sqrt{7}}{3} \approx 3.92$ cm. Maximum volume $V = \frac{1000(10+7\sqrt{7})}{27} \approx 1056$ cm$^3$ [1].
Q7 (4 marks): $P(x) = R(x) - C(x) = 80x - (0.5x^2 + 20x + 400) = -0.5x^2 + 60x - 400$ [1]. $P'(x) = -x + 60 = 0 \Rightarrow x = 60$. $P''(x) = -1 < 0$, so profit is maximised at $60$ units [1]. Maximum profit $= P(60) = \$1400$ [1]. If at most $50$ units can be produced, check the endpoint: $P(50) = \$1350$, which is the maximum possible profit in this domain [1].
Q8 (4 marks): $f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1) = 0$ [1]. Stationary points at $x = 2$ and $x = -1$, both inside $[-2,3]$ [1]. Evaluate: $f(-2) = 1$, $f(-1) = 12$, $f(2) = -15$, $f(3) = -4$ [1]. Maximum value is $12$ (at $x = -1$) and minimum value is $-15$ (at $x = 2$) [1].
Optimisation Problems
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