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.