From Paper 3 to your IA: one way of thinking, two assessments

\(V = x(30-2x)^2\) for \(0 < x < 15\). The graph has its maximum at \(x = 5\), where \(V = 5 \times 20 \times 20 = 2000\ \text{cm}^3\).

IA link: say why you started with a square: it is the simplest case and has only one variable. Show the graph with labelled axes and units, and state the domain. That is criterion A and B work, and it shows the reader your choices (C).

Conjecture: for a square sheet of side \(a\), the best cut is \(x = \dfrac{a}{6}\).

IA link: a table followed by a conjecture written in words and in symbols is exactly how a reader likes to meet a pattern. Choose your cases so the pattern is easy to see, and say why you chose them.

\(V = x(a-2x)^2\), so \(\dfrac{dV}{dx} = (a-2x)^2 - 4x(a-2x) = (a-2x)(a-6x)\).

\(\dfrac{dV}{dx} = 0\) when \(x = \dfrac{a}{6}\) or \(x = \dfrac{a}{2}\). At \(x = \dfrac{a}{2}\) the box has no base and \(V = 0\), so that is not the maximum. The gradient changes from positive to negative at \(x = \dfrac{a}{6}\), so it is a maximum.

\(V_{\max} = \dfrac{a}{6}\left(\dfrac{2a}{3}\right)^2 = \dfrac{2a^3}{27}\). Check: \(a = 30\) gives \(2000\).

IA link: this is criterion E: correct calculus, explained, with the end-points dealt with. Checking the general result against the case you started with is a short line that reviewers notice.

\(V = x(20-2x)(30-2x)\). The GDC gives the maximum at \(x \approx 3.92\), with \(V \approx 1056.3\ \text{cm}^3\).

Using \(x = \dfrac{20}{6} \approx 3.33\) gives \(V \approx 1037.0\ \text{cm}^3\), which is less. So the rule from the square does not carry over: the best cut on a rectangle is a bigger fraction of the short side.

IA link: when an idea fails, say so plainly and say what you learned. That is reflection (D), not a weakness. Many strong explorations turn on one moment like this.

\(V = x(a-2x)(b-2x) = 4x^3 - 2(a+b)x^2 + abx\), so \(\dfrac{dV}{dx} = 12x^2 - 4(a+b)x + ab\).

Setting this to zero and using the quadratic formula: \(x = \dfrac{4(a+b) \pm \sqrt{16(a+b)^2 - 48ab}}{24} = \dfrac{(a+b) \pm \sqrt{a^2 - ab + b^2}}{6}\).

The cut must satisfy \(x < \dfrac{a}{2}\). Since \(b \ge a\), \(b^2 - ab \ge 0\), so \(\sqrt{a^2 - ab + b^2} \ge a\) and the plus root is at least \(\dfrac{2a + b}{6} \ge \dfrac{a}{2}\): outside the domain. So the best cut is the minus root. For \(20 \times 30\) it gives \(3.92\), matching (d).

IA link: going from numbers to letters is the "abstraction" step the new criteria name directly. Always check the general formula against a case you already know: put \(a = b\) and it gives \(\dfrac{a}{6}\).

For \(b = 60, 200, 2000\) the best \(x\) is about \(4.51, 4.87, 4.99\). It approaches \(5 = \dfrac{a}{4}\).

Reason: a very long sheet is nearly a gutter. Its volume is close to (cross-section) \(\times\) length, and the cross-section \(x(a-2x)\) is largest at \(x = \dfrac{a}{4}\). So the answer moves from \(\dfrac{a}{6}\) (square) towards \(\dfrac{a}{4}\) (very long strip).

IA link: a limit with a physical meaning is a strong way to finish. It shows you understand the result, not just that you found it (D, and interpretation from 2029).

Every number from 1 to 20 is a staircase number except 1, 2, 4, 8 and 16. For example \(3 = 1+2\), \(6 = 1+2+3\), \(10 = 1+2+3+4\), \(12 = 3+4+5\), \(15 = 7+8 = 4+5+6 = 1+2+3+4+5\), \(20 = 2+3+4+5+6\).

IA link: work systematically (one number at a time, shortest staircase first) and say that you did. A reader trusts a pattern more when they can see how the evidence was collected.

The powers of 2: \(1, 2, 4, 8, 16, \dots\) Every other positive integer is a staircase number.

IA link: a conjecture is worth more if you test it on a case you did not use to make it. Check 32 and 30 before trying to prove anything.

The terms are \(s, s+1, \dots, s+k-1\): an arithmetic series with \(k\) terms, first term \(s\) and last term \(s+k-1\). So \(N = \dfrac{k}{2}\left(s + s + k - 1\right)\), which gives \(2N = k(2s + k - 1)\).

IA link: one good piece of notation turns lots of examples into one equation. Define every letter before you use it (criterion B).

The factors \(k\) and \(2s + k - 1\) differ by \(2s - 1\), which is odd, so one of them is odd and the other is even. Both are at least 2, and the odd one is at least 3.

So \(2N\) has an odd factor of at least 3. A power of 2 has no odd factor greater than 1, and neither does twice a power of 2. So \(N\) is not a power of 2.

IA link: parity arguments are short and convincing. Explain the key idea in a sentence before the algebra, so the reader knows where you are heading.

Write \(2N = d \times \dfrac{2N}{d}\). Here \(d\) is odd and \(\dfrac{2N}{d}\) is even, and both are at least 2. Let \(k\) be the smaller of the two and \(L\) the larger. Since one is odd and one is even, \(L - k\) is odd, so \(L \ne k\).

Put \(s = \dfrac{L - k + 1}{2}\). This is a positive integer, because \(L - k\) is odd and positive. Then \(2s + k - 1 = L\), so \(k(2s + k - 1) = 2N\): a staircase of \(k\) terms starting at \(s\).

Check with \(N = 20\), \(d = 5\): \(2N = 5 \times 8\), \(k = 5\), \(L = 8\), \(s = 2\), giving \(2+3+4+5+6\).

IA link: proving both directions is the difference between "I think" and "I know". Say clearly which direction each part proves.

15 has 3, 30 has 3, 45 has 5 and 105 has 7. In each case that is one less than the number of odd divisors: 15 has 4 odd divisors (1, 3, 5, 15), 30 has 4, 45 has 6 and 105 has 8.

Conjecture: the number of staircases equals (number of odd divisors of \(N\)) \(- 1\). The proof in (e) suggests why: each odd divisor \(d \ge 3\) gives a different staircase. We checked the conjecture by computer for every \(N\) up to 500.

IA link: say what you have shown and what you have only checked. "Verified by computer up to 500" is honest and is a good limitation to discuss.