Maximum Volume of an Open Box

The open-box problem is a classic 0606 optimisation: cut squares from the corners of a sheet, fold up the sides, and maximise the volume. The twist is a rejected root, one solution makes a side zero, so no box exists.

An open box is made from a square sheet of side $12$ cm by cutting a square of side $x$ cm from each corner and folding up the sides. Show that the volume is $V = x(12 - 2x)^2$, and find the value of $x$ that maximises it. [7]

The working

Step 1, form $V$. The base is $(12 - 2x)$ square, the height is $x$: $V = x(12 - 2x)^2 \quad \text{(M1, answer shown)}$

Step 2, expand ready to differentiate: $V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3 \quad \text{(M1)}$

Step 3, differentiate and set to zero: $\frac{dV}{dx} = 144 - 96x + 12x^2 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6) = 0 \quad \text{(M1, A1)}$ $x = 2 \quad \text{or} \quad x = 6$

Step 4, reject the impossible root. $x = 6$ makes the base $12 - 2(6) = 0$, no box, so reject it. Thus $x = 2$. (M1)

Step 5, confirm a maximum with the second derivative: $\frac{d^2V}{dx^2} = -96 + 24x; \quad \text{at } x = 2: \ -96 + 48 = -48 < 0 \Rightarrow \text{maximum} \quad \text{(A1, A1)}$

So $x = 2$ cm maximises the volume (giving $V = 2(8)^2 = 128\ \text{cm}^3$).

Where the marks are won and lost

• Reject $x = 6$ with a reason (the base would vanish). Carrying it forward, or dropping it silently, costs a mark.
• The second-derivative check confirms $x = 2$ is a maximum, its own mark.
• Expanding $(12 - 2x)^2 = 144 - 48x + 4x^2$ correctly (with the middle term) is essential before differentiating.

Common mistakes

• Keeping $x = 6$ as a valid answer.
• Expanding $(12 - 2x)^2$ without the middle term.
• Skipping the maximum-justification step.

Full method: Stationary Points notes. Topic home: Calculus pillar.