Mathematical Methods · Units 3 & 4

Rates of Change and Optimisation

Master average versus instantaneous rate of change and solve maximum and minimum problems with calculus, the easy way. Plain English intuition, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Speed is the perfect place to start, because everyone already knows the difference without naming it. Your average speed on a trip is the whole distance divided by the whole time. Your speed right now, the number on the speedometer, is something else entirely. Calculus is what lets you flip between these two. Once you can, you can also answer the question every engineer, farmer and business asks: what is the best I can possibly do? That second question is optimisation, and it is just the same idea pushed one step further.

The whole topic in three chunks

1 · Average vs instantaneous rate

Average rate $= \dfrac{f(b)-f(a)}{b-a}$ , the gradient of the chord between two points
Instantaneous rate $= f'(a)$ , the gradient of the tangent at one point
Average uses the original $f$ ; instantaneous needs the derivative first

2 · Set up and solve the optimisation

Write the quantity as one function of one variable (use a constraint to remove the other)
Differentiate and set $\dfrac{dy}{dx} = 0$ to find the stationary points

3 · Confirm and check the edges

Test each turning point with the second derivative: $f''<0$ max, $f''>0$ min
On a closed interval, also test both endpoints, then answer the exact question with units

Two kinds of rate, and why they differ

A curve that bends is changing its steepness all the time, so “how fast is it changing” has two honest answers depending on whether you zoom out or zoom in.

The average rate of change of $f$ between $x = a$ and $x = b$ is the rise over the run across the whole gap. It is the gradient of the straight line joining the two endpoints.

\text{average rate} = \frac{f(b) - f(a)}{b - a}

The instantaneous rate of change at a single point is the derivative evaluated there. It is the steepness of the curve at exactly that spot.

\text{instantaneous rate at } x = a \text{ is } f'(a)

The order of operations matters. For the average rate you plug numbers into the original function $f$ . For the instantaneous rate you differentiate first and then substitute. Mixing these up is one of the most common reported slips.

For h(t) = 12t − t², the red secant from t=0 to t=4 has gradient 8 (the average rate), while the blue tangent at t=4 has gradient 4 (the instantaneous rate).

The classic trap: average value is not average rate

Examiners flag this every single year, so it is worth pinning down. There are two different things that both contain the word “average”.

Average rate of change measures how fast the quantity is changing, which is $\dfrac{f(b)-f(a)}{b-a}$ .
Average value of the function measures how big the quantity is on average, which is $\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$ .

They answer different questions and almost never give the same number. If the word “rate” appears, you want the first one.

Setting up an optimisation problem

Optimisation means finding the biggest or smallest value something can take. The maths is short. The hard part is turning a word problem into one clean function, so do that carefully first.

Name your variable and write the quantity you want to optimise in terms of it.
If two variables appear, use a constraint from the question to remove one, so you are left with a function of a single variable.
Differentiate and set the derivative to zero. Solving gives the candidate turning points.
Decide whether each turning point is a maximum or a minimum, then answer the exact question asked.

That third step is the heart of it. At a smooth peak or valley the curve is momentarily flat, so its gradient is zero. A maximum or minimum sits where the derivative is zero.

\frac{dy}{dx} = 0

Learn it, then fade the help

1 · I do — fully worked

Maximise a fenced pen. A rectangular pen is built against a straight wall using $40$ m of fencing for the three open sides. The side perpendicular to the wall is $x$ m, so the two perpendicular sides use $2x$ m and the side parallel to the wall is $40 - 2x$ . The area is $A(x) = x(40 - 2x) = 40x - 2x^2$ . Differentiate: $A'(x) = 40 - 4x$ . Set $A'(x) = 0$ : $40 - 4x = 0 \implies x = 10$ . Confirm a maximum: $A''(x) = -4 < 0$ (concave down). So $A(10) = 10(40 - 20) = 200$ , a maximum area of $200\ \text{m}^2$ .

2 · We do — your turn, guided

Same setup, your turn. A pen against a wall uses $60$ m of fencing for the three open sides, with the perpendicular side $x$ m. The area is $A(x) = x(60 - 2x)$ . Differentiate to get $A'(x) = 60 - \;?\;$ , then set it to zero and solve for $x$ .

Check your step

$A'(x) = 60 - 4x$ . Setting $60 - 4x = 0$ gives $x = 15$ . Since $A''(x) = -4 < 0$ it is a maximum, and $A(15) = 15(60 - 30) = 450\ \text{m}^2$ .

3 · You do — solo

No scaffolding now. A pen against a wall uses $24$ m of fencing for the three open sides, with the perpendicular side $x$ m. Find the value of $x$ that maximises the area, and state that area.

Reveal solution

$A(x) = x(24 - 2x) = 24x - 2x^2$ , so $A'(x) = 24 - 4x$ . Setting $A'(x) = 0$ gives $x = 6$ . As $A''(x) = -4 < 0$ this is a maximum, and $A(6) = 6(24 - 12) = 72\ \text{m}^2$ .

Confirming a maximum or a minimum

Finding where the derivative is zero is not the end. You still need to know which kind of turning point you have. The quickest check is the second derivative.

If $f'(x_0) = 0$ , then:

$f''(x_0) < 0$ means the curve is concave down, so you have a maximum.
$f''(x_0) > 0$ means the curve is concave up, so you have a minimum.

A sign diagram of $f'$ either side of the point works just as well and is handy when the second derivative is awkward.

Do not forget the endpoints

Here is the mistake that quietly costs marks. On a closed interval such as $[0, 6]$ , the largest or smallest value does not have to happen at a turning point. It can happen right at the edge of the allowed domain.

So when the domain is restricted, your list of candidates is every stationary point plus both endpoints. Evaluate the original function at each, then compare. The winner is whichever gives the biggest or smallest value.

Saying the answer properly

The calculus can be perfect and still lose a mark on presentation. Reports note the same fixable habits over and over. If a question asks for the maximum value, state that value, not just the $x$ that produced it. If it asks for the point, give both coordinates inside round brackets. Keep an exact answer exact unless the question says round. And always sanity check that your answer sits inside the allowed domain.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Write the average rate of change of

f

between

x=a

and

x=b

What is the instantaneous rate of change at

x = a

, and how is it found?

What condition does the derivative satisfy at a maximum or minimum?

After

f'(x_0) = 0

, how does the second derivative tell max from min?

On a closed interval, which points must you test for the absolute max or min?

How do you find the maximum rate of change of a quantity

N(t)

Recall · Stationary Points

What kinds of stationary point can occur where

f'(x) = 0

Recall · The Definite Integral and Area

How does average value differ from average rate of change?

Work through the recipe in the Worked Examples tab, then test yourself in Try It.

Worked Example 1Maximising a triangular area, from a real exam

A theme park is planned whose boundaries form the triangle $\triangle OAB$ , where $O$ is the origin, $A$ is at $(k, 0)$ and $B$ is at $(k, g(k))$ , with $k \in (0, 4)$ and $g(x) = 12x - 3x^2$ . Find the maximum possible area of the theme park, in $\text{km}^2$ .

1
The triangle has base $OA = k$ and height $AB = g(k) = 12k - 3k^2$ , so the area is $A(k) = \tfrac12 k\,g(k)$ .
$A(k) = \frac{1}{2}k(12k - 3k^2) = 6k^2 - \frac{3}{2}k^3$
2
Differentiate the area function and set it to zero to find the turning point.
$A'(k) = 12k - \frac{9}{2}k^2 = 0 \implies k\left(12 - \frac{9}{2}k\right) = 0 \implies k = \frac{8}{3}$
3
Substitute $k = \tfrac{8}{3}$ back into $A(k)$ . The examiner report flags writing the derivative in the wrong variable and arithmetic slips here as the common errors.
$A\!\left(\tfrac{8}{3}\right) = 6\cdot\tfrac{64}{9} - \tfrac{3}{2}\cdot\tfrac{512}{27} = \tfrac{128}{9}$

Answer

\text{maximum area} = \tfrac{128}{9} \text{ km}^2

VCAA 2023 Mathematical Methods Exam 1, Q9c

Worked Example 2Average rate of change of temperature, from a real exam

For a temperature model $f$ , where $f(0) = 12$ and $f\left(\tfrac12\right) = 22$ , find the average rate of change in temperature predicted by the model between $t = 0$ and $t = \tfrac12$ . Give your answer in degrees Celsius per hour.

1
Average rate of change is the change in temperature over the change in time. Use the original function values at the endpoints, not the derivative.
$\frac{f\!\left(\tfrac12\right) - f(0)}{\tfrac12 - 0} = \frac{22 - 12}{\tfrac12 - 0}$
2
Evaluate. The report notes a common error was substituting $t=\tfrac12$ into $12 + 30t$ (giving $27$ , then $30$ ), or finding the average value instead of the average rate.
$\frac{10}{0.5} = 20$

Answer

\text{average rate of change} = 20 \text{ }^{\circ}\text{C/h}

VCAA 2024 Mathematical Methods Exam 2, Section B Q2b

Worked Example 3Average versus instantaneous

A drone rises so its height in metres after $t$ seconds is $h(t) = 12t - t^2$ for $0 \le t \le 12$ . Find the average rate of change of height over the first $4$ seconds, then the instantaneous rate at $t=4$ .

1
Average rate of change is the change in height over the change in time. Work out $h(4)$ and $h(0)$ first.
$h(4) = 12(4) - 4^2 = 32, \quad h(0) = 0$
2
Divide the rise by the time taken. This is the gradient of the straight line joining the two points.
$\frac{h(4) - h(0)}{4 - 0} = \frac{32}{4} = 8$
3
Instantaneous rate is the derivative. Differentiate $h(t)$ .
$h'(t) = 12 - 2t$
4
Substitute $t=4$ into the derivative, not into $h$ .
$h'(4) = 12 - 2(4) = 4$

Answer

\text{average rate} = 8 \text{ m/s}, \quad \text{instantaneous rate at } t=4 \text{ is } 4 \text{ m/s}

Worked Example 4Maximising an area

A rectangular pen is built against a straight wall using $40$ metres of fencing for the three open sides. If the side perpendicular to the wall is $x$ metres, find the value of $x$ that gives the largest area, and state that area.

1
Two sides of length $x$ use up $2x$ metres, so the side parallel to the wall is $40 - 2x$ .
$A(x) = x(40 - 2x) = 40x - 2x^2$
2
Differentiate the area function.
$A'(x) = 40 - 4x$
3
Set the derivative to zero to find the turning point, then solve for $x$ .
$40 - 4x = 0 \implies x = 10$
4
Confirm it is a maximum: $A''(x) = -4 < 0$ , so the curve is concave down. Then find the area.
$A(10) = 10(40 - 20) = 200$

Answer

x = 10 \text{ m gives a maximum area of } 200 \text{ m}^2

Worked Example 5Maximum rate of change

The number of bacteria in a dish is $N(t) = 60t^2 - t^3$ thousand, for $0 \le t \le 40$ . Find the time at which the bacteria are growing fastest, and the maximum instantaneous rate of growth.

1
The growth rate is the derivative $N'(t)$ . We want the time where this rate is largest, so we differentiate again and set $N''(t)=0$ .
$N'(t) = 120t - 3t^2$
2
To maximise the rate, set the derivative of the rate to zero.
$N''(t) = 120 - 6t = 0 \implies t = 20$
3
The maximum rate is the value of $N'(t)$ at $t=20$ , not the value of $N(t)$ .
$N'(20) = 120(20) - 3(20)^2 = 2400 - 1200 = 1200$

Answer

t = 20 \text{, maximum instantaneous rate} = 1200 \text{ thousand per unit time}

Practice test

Try it yourself

9 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A trapezium has parallel sides of length $x$ (top) and $3x$ (bottom), with two equal slant sides of length $10$ and a horizontal offset of $x$ on each side. The value of $x$ which maximises the area of the trapezium is:

1mark

VCAA 2024 Mathematical Methods Exam 2, Section A Q18

Q2.A cylinder of height $h$ and radius $r$ is formed from a thin rectangular sheet of metal of length $x$ and width $y$ , by cutting along the dashed lines shown. The volume of the cylinder, in terms of $x$ and $y$ , is given by:

1mark

VCAA 2023 Mathematical Methods Exam 2, Section A Q17

Q3.The chart shows the daily price of a stock market share over a $30$ -day period. Over which of the following time intervals did the daily price undergo the greatest average rate of change?

1mark

VCAA 2025 Mathematical Methods Exam 2, Section A Q11

Q4.For $f(x) = x^2 - 6x$ , the average rate of change between $x=1$ and $x=4$ is:

1mark

Need a hint?

Average rate is rise over run. Use the original function: compute

f(4)

and

f(1)

, then divide the difference by

4 - 1

Q5.The instantaneous rate of change of $g(x) = 2x^3 - 5x$ at $x=2$ is:

1mark

Need a hint?

Instantaneous rate means differentiate first, then substitute. Find

g'(x)

and only afterwards put

x=2

in.

Q6.A box has volume $V(x) = x(10 - 2x)^2$ for $0 < x < 5$ . The value of $x$ that maximises the volume is:

1mark

Need a hint?

Expand first, differentiate, then set

V'(x)=0

. Discard any solution that lands on the domain boundary where the volume is zero.

Q7.On $[0, 6]$ , the function $h(x) = x^3 - 9x^2 + 24x$ has its absolute maximum at:

1mark

Need a hint?

On a closed interval the winner can be an endpoint. Test every stationary point and both endpoints, then compare the values.

Q8.The temperature is $T(t) = 18 + 6t - t^2$ degrees, $0 \le t \le 6$ . The maximum instantaneous rate of change of temperature occurs at:

1mark

Need a hint?

You are maximising the rate

T'(t)

, not

T

itself. Look at where

T'(t)

is largest across the interval, including the endpoints.

Q9.A farmer encloses a rectangular paddock of area $50$ square metres. The length is $x$ metres and the width is $\dfrac{50}{x}$ metres. Find the value of $x > 0$ that minimises the perimeter $P(x) = 2x + \dfrac{100}{x}$ , and state the minimum perimeter. Show every step.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What is the difference between average and instantaneous rate of change?

Average rate of change is the overall change spread across an interval, found by dividing the change in the function by the change in x, like your average speed on a whole trip. Instantaneous rate is how fast it is changing at a single moment, found from the derivative at that point, like the number on the speedometer right now.

When I do an optimisation problem, how do I know if I have a maximum or a minimum?

After setting the derivative to zero, check the second derivative at that point. If it is negative the curve is concave down and you have a maximum, and if it is positive the curve is concave up and you have a minimum. A sign diagram of the first derivative either side of the point works just as well.

Why do I have to check the endpoints in optimisation questions?

On a closed interval the largest or smallest value does not always happen at a turning point. It can happen right at the edge of the allowed domain. So your candidates are every stationary point plus both endpoints, and you compare the function values at all of them to find the true maximum or minimum.

Rates of Change and Optimisation

Learn

Two kinds of rate, and why they differ

The classic trap: average value is not average rate

Setting up an optimisation problem

Confirming a maximum or a minimum

Do not forget the endpoints

Saying the answer properly

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions