A school compares two medal suppliers for sports day. Company M charges a \(\pounds35\) set-up fee plus \(\pounds3\) per medal. Company N charges \(\dfrac{x...

Writing both companies' costs in terms of \(x\) and subtracting one from the other, as a single fraction, allows direct comparison at specific group sizes and reveals the crossover point where the two companies charge the same; beyond that point, the squared term in Company N's formula eventually dominates.

(a) Company M's cost is the set-up fee plus \(\pounds3\) per medal: \(3x+35\). [1 mark]

(b) \(N-M = \left(\dfrac{x^{2}}{4}+2x\right)-(3x+35) = \dfrac{x^{2}}{4}-x-35\). Writing this as a single fraction over 4: \(\dfrac{x^{2}-4x-140}{4}\). [2 marks]

(c) At \(x=12\): \(N-M = \dfrac{144-48-140}{4} = \dfrac{-44}{4} = -11\). Since this is negative, Company N is cheaper, by \(\pounds11\). [2 marks]

(d) At \(x=16\): \(N-M = \dfrac{256-64-140}{4} = \dfrac{52}{4} = 13\). Since this is positive, Company M is cheaper, by \(\pounds13\). [1 mark]

(e) Setting \(N-M=0\): \(x^{2}-4x-140=0\), which factorises as \((x-14)(x+10)=0\), giving \(x=14\) or \(x=-10\). Since \(x\) must be positive, \(x=14\) medals. [2 marks]

(f) For \(x\gt14\), the \(\dfrac{x^{2}}{4}\) term in Company N's cost grows much faster than Company M's purely linear \(3x\) term, so \(N-M\) becomes positive and keeps increasing. This means Company M gives better value once large numbers of medals are ordered. [1 mark]

Exam tip: whenever one cost formula contains \(x^{2}\) and another is purely linear in \(x\), the quadratic one will always eventually overtake the linear one for large enough \(x\) - the crossover point found by solving \(N-M=0\) marks exactly where that happens.