Optimisation in Calculus | IEB Grade 12 Maths

Optimisation uses the derivative to find the largest or smallest possible value of a quantity, such as the biggest area for a fixed fence or the least material for a fixed volume. It is a favourite IEB problem-solving question in Paper 1, often 6 to 10 marks.

Maximum and minimum problems

Every optimisation problem follows the same plan:

Name the variables and write down what must be maximised or minimised.
Write a constraint that links the variables.
Substitute so the quantity depends on one variable only.
Differentiate and set equal to 0, then solve.
Confirm it is a maximum or minimum, with the second derivative or the context.
Answer the question asked, with units.

Example. Two positive numbers add up to 20. Find the largest possible product.

Let the numbers be $x$ and $20 - x$ . Then $P = x (20 - x) = 20 x - x^{2}$ .

$P^{'} (x) = 20 - 2 x = 0 \Rightarrow x = 10$

$P^{''} (x) = - 2 < 0$ , so this is a maximum. The largest product is $10 \times 10 = 100$ .

Surface area and volume problems

For three-dimensional problems you must know the right prism formulae. They are not given in the exam. Cone, sphere and pyramid formulae are supplied in the question.

Rectangular prism: $V = l bh$ and $S = 2 (l b + l h + bh)$
Cylinder: $V = π r^{2} h$ and $S = 2 π r^{2} + 2 π r h$ (closed)
Open containers lose one face, so check the wording.

Example. A closed box with a square base of side $x$ cm must hold 4 000 cm³. Minimise the surface area.

Constraint: $x^{2} h = 4000$ , so $h = \frac{4000}{x ^{2}}$ .

$S = 2 x^{2} + 4 x h = 2 x^{2} + \frac{16000}{x}$

$S^{'} (x) = 4 x - \frac{16000}{x ^{2}} = 0 \Rightarrow x^{3} = 4000 \Rightarrow x \approx 15, 87 cm$

Then $h = x$ , so the best box is a cube. $S^{''} (x) = 4 + \frac{32000}{x ^{3}} > 0$ confirms a minimum.

Key terms

Constraint: A fixed condition that links the variables, such as a set volume or length of fencing.
Objective function: The quantity you maximise or minimise, written in terms of one variable.
Second derivative test: $f^{''} (x) < 0$ at a stationary point means a maximum; $f^{''} (x) > 0$ means a minimum.

Common mistakes

Differentiating before the quantity is written in terms of a single variable.
Giving the $x$ -value as the answer when the question asks for the maximum area or minimum cost.
Using a closed-box surface area formula for an open box.
Not showing that the stationary point is a maximum or minimum when asked.
Leaving out units.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Two positive numbers have a sum of 20. Determine the numbers if their product is a maximum.

Worked solution

Hidden - try the question first

Solution

Let the numbers be $x$ and $20 - x$ : $P = 20 x - x^{2}$
$P^{'} (x) = 20 - 2 x = 0$
$x = 10$
The numbers are 10 and 10

Question 2

6 marks total

routine

A farmer has 240 m of fencing to fence a rectangular camp next to a straight river. No fence is needed along the river.

Let the two sides perpendicular to the river each be $x$ m. Determine the maximum area of the camp.

Worked solution

Hidden - try the question first

Solution

Side parallel to the river: $x$
$A = x (240 - 2 x) = 240 x - 2 x^{2}$
$A^{'} (x) = 240 - 4 x = 0$
$x = 60$ m
$A^{''} (x) = - 4 < 0$ , so a maximum
$A = 60 \times 120 = 7200$ m²

Question 3

8 marks total

complex

A closed box has a square base of side $x$ cm and a height of $h$ cm. Its volume is 4 000 cm³.

(a) Show that the surface area is $S = 2 x^{2} + \frac{16 000}{x}$ . (3)

(b) Determine the value of $x$ that gives the minimum surface area, and calculate this minimum surface area. Give answers to one decimal place. (5)

Worked solution

Hidden - try the question first

Solution

(a) $x$
$S = 2 x^{2} + 4 x h$
$= 2 x^{2} + 4 x \cdot \frac{4000}{x ^{2}} = 2 x^{2} + \frac{16000}{x}$
(b) $S^{'} (x) = 4 x - \frac{16000}{x ^{2}} = 0$
$x^{3} = 4000$
$x = 34000 \approx 15, 9$ cm
$h = \frac{4000}{x ^{2}} = x$ , so $S = 6 x^{2} = 6 (4000)^{\frac{2}{3}}$
$S \approx 1511, 9$ cm²

Question 4

9 marks total

problem-solving

A company makes cylindrical cups that are open at the top. Each cup must hold 500 cm³. The material for the base and side costs the same per cm².

Let the radius be $r$ cm and the height $h$ cm.

(a) Show that the area of material used is $A = π r^{2} + \frac{1000}{r}$ . (3)

(b) Determine the radius that uses the least material. Give your answer to two decimal places. (4)

Worked solution

Hidden - try the question first

Solution

(a) $x$
Open top: $A = π r^{2} + 2 π r h$
$= π r^{2} + 2 π r \cdot \frac{500}{π r ^{2}} = π r^{2} + \frac{1000}{r}$
(b) $A^{'} (r) = 2 π r - \frac{1000}{r ^{2}} = 0$
$r^{3} = \frac{500}{π}$
$r = 3 \frac{500}{π} \approx 5, 42$ cm
$A^{''} (r) = 2 π + \frac{2000}{r ^{3}} > 0$ , so this is a minimum
(c) $h = \frac{500}{π r ^{2}} = \frac{500}{π r ^{3}} \cdot r$
and $π r^{3} = 500$ , so $h = r$

FAQs

Do I need to know volume formulae for IEB optimisation questions?

You must know right prism and cylinder formulae. Formulae for cones, spheres and pyramids are given in the question.

Do I have to prove that my answer is a maximum or minimum?

If the question asks for it, yes. Even when it does not, a one-line second derivative check is good practice and sometimes earns a mark.

How do I start an optimisation question if I am stuck?

Write down what you are maximising or minimising, then the fixed condition. Use the condition to remove the second variable. Most of the marks are in setting this up.

Optimisation

Maximum and minimum problems

Surface area and volume problems

Key terms

Common mistakes

Practice questions

FAQs

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