Optimisation and rates of change: VCE Math Methods Unit 3

What this dot point is asking

VCAA wants you to apply differentiation to real-world problems where a quantity has to be maximised or minimised, or where the rate at which something is changing must be found. Optimisation is almost guaranteed in Paper 2 Section B and shows up regularly in SACs.

Rates of change

Given a modelled quantity $Q(t)$ as a function of time, the instantaneous rate of change is $\frac{dQ}{dt}$. Read carefully whether the question asks for an average rate ($\frac{Q(b) - Q(a)}{b - a}$) or an instantaneous rate at a specific time ($Q'(t)$).

Sign conventions:

• Positive rate: $Q$ is increasing.
• Negative rate: $Q$ is decreasing.
• Zero rate: $Q$ has a stationary value.

Always state units (e.g. metres per second, dollars per item, degrees per minute) when interpreting the answer in context.

Worked example

A balloon's volume in cubic centimetres at time $t$ seconds is $V(t) = 100 + 30 t - t^3$ for $t \in [0, \sqrt{30}]$.

The rate of change at $t = 2$ is $V'(t) = 30 - 3 t^2$, so $V'(2) = 30 - 12 = 18$ cubic centimetres per second (positive, so the balloon is inflating).

The volume is maximum when $V'(t) = 0$: $3 t^2 = 30$, $t = \sqrt{10} \approx 3.16$ seconds. At that point $V(\sqrt{10}) = 100 + 30 \sqrt{10} - 10 \sqrt{10} = 100 + 20 \sqrt{10}$ cubic centimetres.

Optimisation: the six-step recipe

Optimisation problems all follow the same structure. The recipe:

1. Read the problem and identify the quantity to be optimised. Volume, area, cost, profit, distance. Call it $Q$.
2. Write $Q$ as a function of the variables, using a diagram if helpful. Label every dimension.
3. Use any constraint to reduce $Q$ to a function of one variable. Substitute to eliminate the others.
4. Identify the valid domain based on physical or geometric constraints (lengths must be positive, etc.).
5. Differentiate, set $Q'(x) = 0$, and solve to find candidate stationary points.
6. Classify and check. Use the second derivative test or sign analysis to confirm max or min. Compare with endpoints if the domain is closed. State the final answer with units.

Endpoint check

If the domain is a closed interval $[a, b]$, the global max or min might occur at an endpoint, not at a stationary point. Always compare $Q(a)$, $Q(b)$, and $Q$ at each interior stationary point.

Worked example: minimum surface area

A closed cylindrical can is to hold $1000$ cm$^3$. Find the radius and height that minimise the surface area.

Step 1

Optimise surface area $S$.

Step 2

$S = 2 \pi r^2 + 2 \pi r h$ (two circular ends, plus lateral surface).

Step 3

Constraint: $\pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$.

Substitute: $S(r) = 2 \pi r^2 + 2 \pi r \cdot \frac{1000}{\pi r^2} = 2 \pi r^2 + \frac{2000}{r}$.

Step 4

Domain: $r > 0$.

Step 5

$S'(r) = 4 \pi r - \frac{2000}{r^2}$. Set to zero: $4 \pi r = \frac{2000}{r^2}$, so $r^3 = \frac{500}{\pi}$ and $r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42$ cm.

Step 6

$S''(r) = 4 \pi + \frac{4000}{r^3} > 0$, so this is a minimum. $h = \frac{1000}{\pi r^2} \approx 10.83$ cm.

Notice $h = 2r$ at the minimum, a standard result for a closed cylinder.

Worked example: maximum revenue

A small business sells $x$ units per week at a price $p = 50 - 0.1 x$ dollars. Revenue is $R(x) = x p = x(50 - 0.1 x) = 50 x - 0.1 x^2$.

$R'(x) = 50 - 0.2 x$. Set to zero: $x = 250$.

$R''(x) = -0.2 < 0$, so this is a maximum.

Maximum revenue $= R(250) = 50(250) - 0.1(62500) = 12500 - 6250 = 6250$ dollars.

Examples in context

Example 1. Maximising box volume. A box with a square base of side $x$ and no lid is built from $300 \text{ cm}^2$ of material. The surface area is $x^2 + 4xh = 300$, so $h = \frac{300 - x^2}{4x}$. The volume is $V = x^2 h = \frac{x(300 - x^2)}{4} = 75x - \frac{x^3}{4}$. Then $V'(x) = 75 - \frac{3x^2}{4} = 0$ gives $x^2 = 100$, so $x = 10$ cm. Since $V''(x) = -\frac{3x}{2} < 0$, this is a maximum: $V = 750 - 250 = 500 \text{ cm}^3$.

Example 2. Rate of cooling. A cup of tea cools as $T(t) = 25 + 70e^{-0.04t}$ degrees, $t$ in minutes. The cooling rate is $T'(t) = -2.8e^{-0.04t}$ degrees per minute. At $t = 15$, $T'(15) = -2.8e^{-0.6} = -2.8 \times 0.5488 = -1.54$ degrees per minute, the instantaneous rate of fall.

Try this

Q1. A rectangle of perimeter $24$ cm has one side $x$. Find $x$ that maximises the area. [3 marks]

• Cue. $A = x(12 - x)$; $A'(x) = 12 - 2x = 0 \Rightarrow x = 6$ (a $6 \times 6$ square).

Q2. A particle's position is $s(t) = t^3 - 6t^2 + 9t$ m. Find its velocity at $t = 1$. [3 marks]

• Cue. $s'(t) = 3t^2 - 12t + 9$; $s'(1) = 3 - 12 + 9 = 0$ m/s.

Q3. A population is $P(t) = 1000e^{0.03t}$. Find the rate of growth at $t = 10$. [3 marks]

• Cue. $P'(t) = 30e^{0.03t}$; $P'(10) = 30e^{0.3} = 30 \times 1.3499 = 40.5$ per unit time.