A triangle is formed by the three lines $y = 10 - 2x,{\text{ }}y = mx$ and $y = -\frac{1}{m}x$, where $m > \frac{1}{2}$.

Find the value of m for which the area of the triangle is a minimum.

attempt to find intersections     M1

intersections are $\left( {\frac{{10}}{{m + 2}},\frac{{10m}}{{m + 2}}} \right){\text{ and }}\left( {\frac{{10m}}{{2m - 1}}, - \frac{{10}}{{2m - 1}}} \right)$     A1A1

area of triangle $= \frac{1}{2} \times \frac{{\sqrt {100 + 100{m^2}} }}{{(m + 2)}} \times \frac{{\sqrt {100 + 100{m^2}} }}{{(2m - 1)}}$     M1A1A1

$= \frac{{50(1 + {m^2})}}{{(m + 2)(2m - 1)}}$

minimum when $m = 3$     (M1)A1

[8 marks]

Most candidates had difficulties with this question and did not go beyond the determination of the intersection points of the lines; in a few cases candidates set up the expression of the area, in some cases using unsimplified expressions of the coordinates.