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 $SN = x \; \text{km}$, then $NM =1-x\; \text{km}$. Using Pythogaras' theorem for the right angled triangle $NMF$: $$NF^2 = 1 + (1-x)^2 = x^2 -2x + 2.$$ The total time taken $T$ is given by $$T = {x\over 10} + {(x^2 - 2x + 2)^{1/2}\over 6}$$ By differentiation $${dT\over dx} = {1\over 10} + {2x-2\over 12(x^2 -2x + 2)^{1/2}}.$$ For a maximum or minimum time this derivative is zero, so that |

$$3(x^2 -2x + 2)^{1/2}= 5(1-x)$$ then by squaring both sides,
$$9(x^2 - 2x + 2) = 25(1 - 2x +x^2)$$ which gives $$16x^2 -32x + 7
= 0$$ which factorizes to give $$(4x-1)(4x-7) = 0$$ so $x=1/4$ or
$x=7/4$.

As $x\leq 1$ the critical time is given by $x = 0.25 \;
\text{km} = 250 \; \text{m}$. 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.