The solution below was sent in by Taryn from Kerang Technical High School. The quickest route takes 14 minutes walking exactly one quarter of the way along the edge and then in a straight line across the field. Congratulations to all of you who found this solution. Other good solutions were sent in by Jenny from KJS, by Thomas and by Andrei from Tudor Vianu National College, Romania.

Let SNF be my path to cross the field. Let SN = x km, then NM =1-x km. Using Pythogaras' theorem for the right angled triangle NMF:

{NF}^{2} = 1 + (1 - x)^{2} = x^{2} - 2 x + 2 .

The total time taken

T

is given by

T = \frac{x}{10} + \frac{(x^{2} - 2 x + 2)^{1 / 2}}{6}

By differentiation

\frac{dT}{dx} = \frac{1}{10} + \frac{2 x - 2}{12 (x^{2} - 2 x + 2)^{1 / 2}} .

For a maximum or minimum time this derivative is zero, so that

3 (x^{2} - 2 x + 2)^{1 / 2} = 5 (1 - x),

then by squaring both sides,

9 (x^{2} - 2 x + 2) = 25 (1 - 2 x + x^{2})

which gives

16 x^{2} - 32 x + 7 = 0

which factorizes to give

(4 x - 1) (4 x - 7) = 0

x = 1 / 4

x = 7 / 4

As $x \leq 1$ the critical time is given by $x = 0.25$ km = 250 m. This gives a time of 14 minutes to cross the field. Increasing and decreasing $x$ slightly increases the time taken so this is a minimum time.