# Problem Solving with or without calculus

Thursday 12th November 2020

As part of a recent Network Meeting we looked at the question below which could be used with GCSE students (without calculus), or A level students (with calculus).

We looked at what assumptions students would need to make and what difficulties they may have, and briefly looked at the source of the question. Here are just two possible solutions:

*Without calculus:*

Let \(P\) be point \((x,y)\) and greatest distance from origin be \(D\).

Look for repeated root \(x^2+y^2=D^2\) and \(3x^2+y^2+15x=18\)

Solve simultaneously: \(2x^2+15x+D^2−18=0\)

Use \(b^2−4ac=0\)

\(15^2−(4)(2)(D^2−18)=0\)

\(225+144=8D^2\)

\(D^2=\frac{369}{8}\)

\(D=\frac{3\sqrt82}{4}\)

*With calculus:*

\(D^2=x^2+y^2\) and \(3x^2+y^2+15x=18\)

So \(D^2=18−15x−2x^2\)

Let \(D^2=L\)

\(\frac{dL}{dx}=−15−4x\) with stationary point when \(\frac{dL}{dx}=0, x=−\frac{15}{4}\)

\(\frac{d^2L}{dx^2}=-4\)\(<\)\(0\) so maximum at \(x=-\frac{15}{4}\)

When \(x=−\frac{15}{4}, y=\bar{+}\frac{3\sqrt57}{4}\)

\(D^2=(−\frac{15}{4})^2+(\frac{3\sqrt57}{4})^2\)

\(D=\frac{3\sqrt82}{4}\)

If you would be interested in attending a similar session, you can find out more information and apply for our next online Teacher Network meeting on the AMSP website.

By **Jean Smith**