What Is Optimisation?

Optimisation is about finding the best possible value -- the maximum or minimum -- of a quantity. Using calculus, we can set up a function for the quantity we want to optimise, differentiate it, and find stationary points.

Common optimisation problems ask you to find the maximum area, minimum cost, shortest distance, or greatest volume under given constraints.

Problem-Solving Strategy

1

Draw a diagram if applicable. Label all variables and known quantities.

2

Identify the quantity to optimise and write it as a function.

3

Use constraints to express the function in terms of a single variable.

4

Differentiate and set the derivative equal to zero to find stationary points.

5

Verify using the second derivative test (or check endpoints if restricted domain).

6

Answer the question with correct units and in context.

Common Types of Optimisation Problems

Maximum Area

Given a fixed perimeter, find the dimensions that give the greatest area.

Minimum Material

Given a fixed volume, find dimensions that minimise the surface area (material used).

Maximum Revenue/Profit

Find the price or quantity that maximises revenue or profit.

Minimum Distance

Find the point on a curve closest to a given point.

Key Vocabulary

Optimise

To find the maximum or minimum value of a quantity subject to given conditions or constraints.

Constraint

A condition that limits or relates the variables, such as a fixed perimeter or budget.

Feasible Domain

The range of valid values for the variable, determined by the physical context of the problem.

Global vs Local

A global maximum/minimum is the absolute greatest/least value; a local one is only the best nearby.

Worked Examples

1

A farmer has 100 m of fencing to make a rectangular paddock against a wall. Find the maximum area.

Step 1: Let width = x metres. Then length = 100 - 2x (fencing for two widths and one length).

Step 2: Area A = x(100 - 2x) = 100x - 2x².

Step 3: dA/dx = 100 - 4x. Set to 0: x = 25.

Step 4: d²A/dx² = -4 < 0, confirming maximum.

Answer: Maximum area = 25 × 50 = 1250 m² when width = 25 m and length = 50 m.

2

Find two positive numbers whose sum is 20 and whose product is maximum.

Step 1: Let the numbers be x and (20 - x).

Step 2: Product P = x(20 - x) = 20x - x².

Step 3: dP/dx = 20 - 2x = 0, giving x = 10.

Step 4: d²P/dx² = -2 < 0 (maximum).

Answer: Both numbers are 10, with maximum product = 100.

3

An open-top box is made from a 24 cm × 24 cm sheet by cutting squares of side x from each corner. Find x for maximum volume.

Step 1: Volume V = x(24 - 2x)² for 0 < x < 12.

Step 2: Expand: V = x(576 - 96x + 4x²) = 4x³ - 96x² + 576x.

Step 3: dV/dx = 12x² - 192x + 576 = 12(x² - 16x + 48) = 12(x - 4)(x - 12) = 0.

Step 4: x = 4 (valid, since x = 12 is an endpoint). d²V/dx² at x = 4: 24(4) - 192 = -96 < 0 (max).

Answer: Cut squares of x = 4 cm for max volume = 4(16)² = 1024 cm³.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

What is the first step in solving an optimisation problem?

Differentiate immediately Set the function equal to zero Find the second derivative Draw a diagram and define variables

Question 2

A rectangular field uses 60 m of fencing on 3 sides (one side is a wall). What width x gives maximum area?

x = 15 m x = 10 m x = 20 m x = 30 m

Question 3

How do you verify that a stationary point gives a maximum (not a minimum)?

Check that f'(x) = 0 Check that f''(x) < 0 Check that f''(x) > 0 Check that x > 0

Question 4

Two positive numbers have a sum of 30. Their product P = x(30 - x). What is the maximum product?

900 200 225 150

Question 5

In optimisation on a closed interval [a, b], where can the absolute maximum or minimum occur?

Only at stationary points Only at the endpoints At x = 0 only At stationary points or at the endpoints of the interval

Optimisation Problems