Writing a complete argument

Communication is assessed

Logical development, precision and clarity is one of the four things the performance standard asks for — it is worth marks in its own right.
Outstanding Scholarship additionally asks for convincing communication.
In practice this means your answer should read as an argument, not a pile of computation.

What a complete argument contains

Definitions. "Let $V$ be the volume in cm³ at time $t$ seconds." Undefined symbols cost marks.
A stated method and reason. One short sentence before the algebra.
Conditions checked. Continuity, differentiability, domain restrictions, denominators that could vanish.
Intermediate values, labelled — not just the final number.
A justification of the nature of a result. Finding $f'(x) = 0$ is not finding a maximum until you have shown it is one.
An interpretation in context, in words, at the end.

Precision that markers look for

Keep exact values ( $\pi$ , surds, logs) until the final line, then round once and say what you rounded to.
Use correct notation: $\dfrac{dy}{dx}$ for a derivative of a function of $x$ , $\dfrac{dV}{dt}$ for a rate with respect to time.
Include units wherever the context has them.
State the domain on which your answer is valid if the context restricts it (lengths are positive, time is non-negative).

Worked Example

An open-topped box is made from a square sheet of card of side $a$ cm by cutting a square of side $x$ cm from each corner and folding up the sides.

Find, in terms of $a$ , the value of $x$ that maximises the volume, and justify that it is a maximum.

Answer:

Set up. After cutting corners of side $x$ , the base is a square of side $(a - 2x)$ and the height is $x$ . So the volume is

$V = x(a - 2x)^2$

This is valid for $0 < x < \dfrac{a}{2}$ , since the base side length must be positive.

Differentiate. Expanding first makes the differentiation cleaner:

$V = x(a^2 - 4ax + 4x^2) = a^2x - 4ax^2 + 4x^3$

$\frac{dV}{dx} = a^2 - 8ax + 12x^2$

Find the stationary points. Setting $\dfrac{dV}{dx} = 0$ :

$12x^2 - 8ax + a^2 = 0$

Factorising:

$(6x - a)(2x - a) = 0 \quad \Rightarrow \quad x = \frac{a}{6} \ \text{ or } \ x = \frac{a}{2}$

$x = \dfrac{a}{2}$ is rejected — it lies outside the domain and gives a box of zero base.

Justify it is a maximum. Differentiating again:

$\frac{d^2V}{dx^2} = -8a + 24x$

At $x = \dfrac{a}{6}$ : $\ \dfrac{d^2V}{dx^2} = -8a + 4a = -4a$ , which is negative for $a > 0$ , so this stationary point is a maximum.

The volume is greatest when $x = \dfrac{a}{6}$ .

Interpretation. The optimal cut is always one sixth of the side length, whatever the size of the sheet — the answer is a fixed ratio, not a fixed length. That is why working in terms of $a$ is more informative than plugging in a number.

Writing a complete argument

Covers 93202

Communication is assessed

Logical development, precision and clarity is one of the four things the performance standard asks for — it is worth marks in its own right.
Outstanding Scholarship additionally asks for convincing communication.
In practice this means your answer should read as an argument, not a pile of computation.

What a complete argument contains

Definitions. "Let $V$ be the volume in cm³ at time $t$ seconds." Undefined symbols cost marks.
A stated method and reason. One short sentence before the algebra.
Conditions checked. Continuity, differentiability, domain restrictions, denominators that could vanish.
Intermediate values, labelled — not just the final number.
A justification of the nature of a result. Finding $f'(x) = 0$ is not finding a maximum until you have shown it is one.
An interpretation in context, in words, at the end.

Precision that markers look for

Keep exact values ( $\pi$ , surds, logs) until the final line, then round once and say what you rounded to.
Use correct notation: $\dfrac{dy}{dx}$ for a derivative of a function of $x$ , $\dfrac{dV}{dt}$ for a rate with respect to time.
Include units wherever the context has them.
State the domain on which your answer is valid if the context restricts it (lengths are positive, time is non-negative).

Worked Example

An open-topped box is made from a square sheet of card of side $a$ cm by cutting a square of side $x$ cm from each corner and folding up the sides.

Find, in terms of $a$ , the value of $x$ that maximises the volume, and justify that it is a maximum.

Answer:

Set up. After cutting corners of side $x$ , the base is a square of side $(a - 2x)$ and the height is $x$ . So the volume is

$V = x(a - 2x)^2$

This is valid for $0 < x < \dfrac{a}{2}$ , since the base side length must be positive.

Differentiate. Expanding first makes the differentiation cleaner:

$V = x(a^2 - 4ax + 4x^2) = a^2x - 4ax^2 + 4x^3$

$\frac{dV}{dx} = a^2 - 8ax + 12x^2$

Find the stationary points. Setting $\dfrac{dV}{dx} = 0$ :

$12x^2 - 8ax + a^2 = 0$

Factorising:

$(6x - a)(2x - a) = 0 \quad \Rightarrow \quad x = \frac{a}{6} \ \text{ or } \ x = \frac{a}{2}$

$x = \dfrac{a}{2}$ is rejected — it lies outside the domain and gives a box of zero base.

Justify it is a maximum. Differentiating again:

$\frac{d^2V}{dx^2} = -8a + 24x$

At $x = \dfrac{a}{6}$ : $\ \dfrac{d^2V}{dx^2} = -8a + 4a = -4a$ , which is negative for $a > 0$ , so this stationary point is a maximum.

The volume is greatest when $x = \dfrac{a}{6}$ .