Quick Route
What is the quickest route across a ploughed field when your speed
around the edge is greater?
Age
16 to 18
Challenge level
Problem
You want to get across a square ploughed field from one corner to the opposite corner as quickly as possible. There is a path along one edge of the field but the rest of the field is ploughed right up to the other three edges. On the ploughed land you can walk at 6 km per hour. From where you are standing you can walk on the path along one edge of the field at 10 km per hour, but there are
no paths along the other edges of the field. What is the best route to take?
Suppose the edge of the field is one kilometre in length. What is the shortest time in which you can cross the field to the opposite corner?
Getting Started
If you go along the edge a distance $x$ then head straight for the
far corner can you work out the time taken as a function of $x$?
Student Solutions
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.
Image
|
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.