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Grade 12
CAPS Syllabus
Optimisation

Applications of Differential Calculus

Optimisation problems connect derivatives to real choices – from fuel-efficient driving to designing containers.

Key Takeaways
  • Optimisation asks us to maximise or minimise a quantity using calculus.
  • Always express the quantity in one variable before differentiating.
  • The derivative equal to zero locates stationary points (possible optima).
  • Use or a sign chart to confirm minima or maxima.
  • State your answer with the context: units, interpretations, and any constraints.

Why Optimisation Matters

Optimisation makes calculus tangible and powerful. Every day, we face questions like "What's the best way to do this?" or "How can I maximize that?" Calculus gives us the mathematical tools to answer these questions precisely.

Maximization Problems

Finding the greatest value:

  • Maximum area with limited fencing
  • Maximum profit in business
  • Maximum height of a projectile
  • Maximum volume of a container
Minimization Problems

Finding the smallest value:

  • Minimum cost of materials
  • Minimum fuel consumption
  • Minimum surface area
  • Shortest distance
The Big Idea

At a maximum or minimum, the function momentarily stops changing — it's neither increasing nor decreasing. This is exactly where the derivative equals zero!

Think of climbing a mountain:

  • While climbing up, your height is increasing (derivative positive)
  • At the peak, you're momentarily neither going up nor down (derivative zero)
  • Going down the other side, your height is decreasing (derivative negative)

Key Mathematical Facts:

  • Stationary points (turning points) occur where
  • Use the second derivative test or interval testing to classify the point as a maximum or minimum
  • Always check the domain: lengths, speeds, volumes, and distances must remain positive (and other physical constraints)
  • Sometimes the maximum or minimum occurs at the boundary of the domain, not at a turning point!

Common Misconception:

Not every point where is a maximum or minimum! You must verify using the second derivative test or sign changes. Some points might be points of inflection where the function doesn't actually change from increasing to decreasing (or vice versa).

Problem-Solving Roadmap

1. Understand the context
Sketch, label, and highlight what needs maximising or minimising.
2. Write constraints
Turn words into equations such as perimeters or totals.
3. Create a one-variable function
Use substitution so the objective depends on one letter only.
4. Differentiate and solve
Find where the derivative equals zero (and is defined).
5. Classify the stationary point
Test using the second derivative or a sign chart.
6. Communicate the answer
Quote the value, units, and why it is a maximum/minimum.

South African Stories

Optimisation challenges pop up in every province. Use these story cards to anchor the maths in local, tangible decisions before diving into formal exercises.

Karoo & Northern Cape

N1 fuel planning
Learners model a long-distance trip from Cape Town to Johannesburg, using calculus to pick the most efficient cruising speed on flat, windy roads.
  • Use f(v) = 3/80 v² - 6v + 245 to explain economy zones
  • Link derivative sign to advice for the driver
  • Discuss impact of speed limits and safety

KwaZulu-Natal coast

Durban harbour deck
A verandah wraps around a cottage above the Indian Ocean. Students maximise the deck area with a fixed length of timber railing.
  • Translate geometry into A(x) = 30x - 3x²
  • Debate practical constraints (minimum widths)
  • Sketch plan and interpret derivative signs

Limpopo cooperative

Agri packaging lab
A farming co-op designs a recyclable juice box with a square base and folding top, minimising cardboard to save costs.
  • Convert surface-area constraints into A(x) = 3000/x + 3x²
  • Use the second derivative to justify the minimum
  • Relate dimensions back to manufacturing realities

Graph & Sketch Lab

Pair every word problem with a visual: a labelled diagram plus a graph of the function we are optimising. These dynamic sketches match the way universities expect you to communicate reasoning.

Garden area vs length
Optimum occurs when l = 40 m and w = 80 m (Michael's garden).
Wall side
l = 40 m
w = 80 m
3200 m²
Fence available: 160 m

Area vs length graph

Fuel model graph
Visualise why 80 km/h minimises the litres per 100 km.

Speed

80 km/h

Fuel

5.0 L/100 km

Derivative

0.00

Optimal

Model

f(v) = 3/80 v² - 6v + 245

Fuel use vs speed

Worked Examples

Study these detailed examples to master the optimization process. Each example follows the 6-step roadmap.

Example 1: Fuel Efficiency (Minimization)
A car uses fuel according to where is the travelling speed (km/h) and is litres per 100 km. Find the most economical speed.

1Understand the context

We want to minimize fuel consumption . The domain is realistic speeds: , probably km/h.

2Write constraints

Already in one variable:

3Create a one-variable function

Already done ✓ We're minimizing

4Differentiate and solve

Step 4a: Find the derivative

$$\begin{aligned} f(v) &= \frac{3}{80}v^2 - 6v + 245 \\ f'(v) &= \frac{3}{80}(2v) - 6 \\ &= \frac{3}{40}v - 6 \end{aligned}$$

Step 4b: Set

$$\begin{aligned} \frac{3}{40}v - 6 &= 0 \\ \frac{3}{40}v &= 6 \\ v &= 6 \times \frac{40}{3} \\ v &= 80 \text{ km/h} \end{aligned}$$

5Classify the stationary point

Using the second derivative test:

$$\begin{aligned} f''(v) &= \frac{3}{40} \\ &= 0.075 > 0 \end{aligned}$$

Since , the function is concave up at , confirming a minimum.

6Communicate the answer

Answer:

The most economical cruising speed is 80 km/h. At this speed, the car uses approximately litres per 100 km.

Why the second derivative test works:

If , the function curves upward like a smile 😊 → minimum
If , the function curves downward like a frown ☹️ → maximum

Example 2: Two Positive Numbers (Maximization)
The sum of two positive numbers is 10. The product is the first number multiplied by the square of the second. Find the numbers that maximize this product.

1-2Understand & Write constraints

Let the two numbers be and .

Constraint:

$$a + b = 10$$

Product to maximize:

$$P = a \cdot b^2$$

3Create one-variable function

From , we get

Substitute into the product:

$$\begin{aligned} P(a) &= a \cdot b^2 \\ &= a \cdot (10 - a)^2 \\ &= a(100 - 20a + a^2) \\ &= 100a - 20a^2 + a^3 \end{aligned}$$

4Differentiate and solve

Differentiate:

$$\begin{aligned} P(a) &= a^3 - 20a^2 + 100a \\ P'(a) &= 3a^2 - 40a + 100 \end{aligned}$$

Set and solve:

$$\begin{aligned} 3a^2 - 40a + 100 &= 0 \end{aligned}$$

Using the quadratic formula :

$$\begin{aligned} a &= \frac{40 \pm \sqrt{1600 - 1200}}{6} \\ &= \frac{40 \pm \sqrt{400}}{6} \\ &= \frac{40 \pm 20}{6} \\ a &= 10 \text{ or } a = \frac{10}{3} \end{aligned}$$

If , then , which gives . This is clearly not a maximum!
So we use , giving

5Classify using second derivative

$$P''(a) = 6a - 40$$

At :

$$P''\left(\frac{10}{3}\right) = 6\left(\frac{10}{3}\right) - 40 = 20 - 40 = -20 < 0$$

Since , we have a maximum ✓

6State the answer

Answer:

The numbers are and .
The maximum product is

Example 3: Garden Against a Wall (Area Maximization)
Michael wants to fence a rectangular vegetable garden against a wall. He has 160 m of fencing. What dimensions maximize the area?

1Understand & sketch

WALL

length = l

(parallel to wall)

width = w

Fence: w + l + w = w + 2l (only 3 sides)

Goal: Maximize area
Constraint: Total fencing m

2-3Use constraint to get one variable

From :

$$w = 160 - 2l$$

Substitute into area formula:

$$\begin{aligned} A &= l \times w \\ A(l) &= l(160 - 2l) \\ &= 160l - 2l^2 \end{aligned}$$

4Differentiate and solve

$$\begin{aligned} A(l) &= 160l - 2l^2 \\ A'(l) &= 160 - 4l \end{aligned}$$

Set :

$$\begin{aligned} 160 - 4l &= 0 \\ 4l &= 160 \\ l &= 40 \text{ m} \end{aligned}$$

Then m

5Verify it's a maximum

$$A''(l) = -4 < 0$$

The second derivative is always negative, confirming the function is concave down everywhere. Therefore, gives a maximum ✓

6Final answer with units

Answer:

Dimensions: Length m (parallel to wall), Width m
Maximum area: m²

Notice the pattern:

For a fixed perimeter, rectangles have maximum area when dimensions are in a 1:2 ratio. Here, 40:80 = 1:2!

Interactive Playground

Interactive Playground
Explore two Grade 12 optimisation staples by moving the sliders.

Length along the wall (l)

Use w + 2l = 160 so w = 160 - 2l.

40 m
Width (w)

80 m

Only three sides are fenced.

Area

3200 m²

100% = 3200 m²

Derivative

0 (A'(l) = 160 - 4l)

optimal

Maximum area reached when and .

Wall side
l = 40 m
w = 80 m
3200 m²
Fence available: 160 m

Area vs length graph

Practice Questions

Additional Practice Questions
Access more practice problems with PDF export functionality. Download questions-only or with complete solutions for offline study.
View Practice Page

Apply what you've learned with these optimization problems. Start with basic questions and progress to more challenging ones.

Question 1
Question 2
Question 3

Self-Check Quiz

Test your understanding of optimization concepts. Try to answer without looking back!

Question 1
What is the first calculus step when solving an optimization problem?
Question 2
If $f'(x) = 0$ and $f''(x) < 0$, what does this tell us about the stationary point?
Question 3
For the garden problem where one side is against a wall with constraint $w + 2l = 160$, which expression for area is correct?
Question 4
A function satisfies $V'(x) = 0$ at $x = 10$ with $V''(10) = -18$. What statement is true?
Question 5
Which sentence best explains why we must express the objective function in terms of a single variable?
Question 6
To find where a parabola $f(x) = x^2$ is closest to a point, you should minimize:
Question 7
A rectangle has fixed perimeter 40 m. What shape maximizes the area?
Question 8
In an optimization problem, if you find $f'(x) = 0$ at $x = 5$, what should you do next?
Question 9
When finding dimensions that maximize volume subject to a constraint, you typically:
Question 10
The second derivative $f''(x) > 0$ at a critical point indicates:

Keep exploring CAPS calculus

Link this optimisation lesson with differential rules, tangents, and rates of change to build a complete Grade 12 toolkit.

Revise differentiationStudy turning points