A-Level Maths: Core Mathematics

Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square. The result changes if, instead of maximising the sum of lengths of sides of the rectangle, we seek to maximise the sum of nth powers of the lengths of those sides for n > 2. What happens if n = 2? What happens if n = 3? Justify your answers.

Let the circle have diameter d and let the length of one side of the rectangle be x and the length of the adjacent side be y.Then, by Pythagoras’s theorem,\[ y = \sqrt{d^2 - x^2}\]and the perimeter P is given by \[P = 2x + 2\sqrt{d^2 - x^2}\]We can find the largest possible value of P as x varies by calculus. We have\[\frac{dP}{dx} = 2 - 2\frac{x}{\sqrt{d^2 - x^2}}\]so for a stationary point, we require (cancelling the factor of 2 and squaring) \[ 1 = \frac{x^2}{d^2 - x^2} i.e. 2x^2 = d^2\]Thus x = d/√2 (ignoring the negative root for obvious reasons). Substituting this into (†) gives y = d/√2, so the rectangle is indeed a square, with perimeter 2√2 d . But is it the maximum perimeter? The easiest way to investigate is to calculate the second derivative, which is easily seen to be negative for all values of x, and in particular when x = d/√2 . The stationary point is certainly a maximum. For the second part, we consider\[f(x) = x^n + (d^2 - x^2)^{n/2}\]The first thing to notice is that f is constant if n = 2, so in this case, the largest (and smallest) value is \(d^2\).For n = 3 , we have\[f'(x) = 3x^2 - 3x(d^2 - x^2)^{1/2}\]so\(f(x)\) is stationary when \(x^4 = x^2(d^2 - x^2)\), i.e. when \(2x^2 = d^2\) as before or when \( x = 0\) Thecorresponding stationary values are \(\sqrt{2d^3}\) and \(2d^3\) so this time the largest value occurs when \(x = 0\)

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins. In how many ways can you make up 20 pence using 20p, 10p, 5p, 2p and 1p coins?

Probably the best approach is to start counting with the arrangements which use as many high denomination coins as possible, then work down.We can make up 10p as follows: 10; 5+5 (one way using two 5p coins); 5+2+2+1, 5+2+1+1+1, 5+1+1+1+1+1, (three ways using one 5p coin); 2+2+2+2+2, 2+2+2+2+1+1, etc, (six ways using no 5p coins); making a total of 11 ways.  We can make up 20p as follows: 20; 10 + any of the 11 arrangements in the first part of the question; 5+5+5+5; 5+5+5+2+2+1, etc (3 ways using three 5p coins); 5+5+2+2+2+2+2, 5+5+2+2+2+2+1+1 etc (6 ways using two 5p coins); 5+2+2+2+2+2+2+2+1, etc (8 ways using one 5 and making 15 out of 2p and 1p coins); 2+2+2+2+2+2+2+2+2+2, etc (11 ways of making 20p with 2p and 1p coins). Grand total = 41.

Suppose that\[3 = \frac{2}{x_1} = x_1 + \frac{2}{x_2} = x_2 + \frac{2}{x_3} = x_3 + \frac{2}{x_4} =  ...\]Guess an expression, in terms of \(n\) , for \(x_n\) . Then, by induction or otherwise, prove the correctness of your guess.

First we put the equations into a more manageable form. Each equality can be written in the form\(3 = x_n + \frac{2}{x_n + 1}\) i.e. \(x_n + 1 = \frac{2}{3-x_n}\)We find \(x_1 = 2/3, x_2 = 6/7, x_3 = 14/15\) and \(x_4 = 30/31\). The denominators give the game away. We guess \[x_n = \frac{2^{n+1} - 2}{2^{n+1} - 1}\]For the induction, we need a starting point: our guess certainly holds for n = 1 (and 2, 3, and 4!). For the inductive step, we suppose our guess also holds for n = k, where k is any integer. If we can show that it then also holds for n = k + 1, we are done. We have, from the equation given in the question\[x_{k+1} = \frac{2}{3-x_k} = \frac{2}{3 - \frac{2^{k+1} - 2}{2^{k+1} - 1}} = \frac{2(2^{k+1} - 1)}{3(2^{k+1} - 1) - (2^{k+1} - 2)} = \frac{2^{k+2} - 2}{2^{k+2} - 1}\]as required

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