1. The Farmer Brown Problem

2. Objectives We are learning to:- - use problems in different ways
- solve problems in many ways
- appreciate other solutions
- be more creative mathematicians
- think beyond one solution
- make connections within mathematics

3. The Problem
When Brownie, the chicken farmer travelled to town at 30km/hr he noticed he arrived an hour too early. He noticed that when he travelled at 20km/hr he arrived an hour too late.

4. The Puzzles How far was the return journey?
How fast should he travel to arrive on time?
How long did it take him to get to town?
How fast should he travel to arrive 2hrs late?
What excuse did he give to Mrs Brown?
What color was his tractor?
eggsactly!

5. Challenge
Solve one
of the
puzzles.

6. A Tablular Solution t v=30 v=20 0 30 0 1 30 20 2 60 40 3 90 60
We are looking for the same distance two hours apart.
Why?
Who would answer with this solution?

7. An Algebraic Solution We know speed = distance/time
So time = distance/speed
And that means…
distance/30 plus 1 hr = distance/20 less 1 hr
WHY?
and solving this gives the ………...

8. A Graphical Solution Distance in 20km Why are there two
120
Why are there two
different starting
points?
Slope = 20
Slope = 30 5
Time in hours

9. Another Graphical Solution
Distance in 20km
120
The slope
of the red line
approximates
the speed.
Prove this!
Slope = 30
Slope = 20 5
Time in hours

10. A Trigonometrical Solution
From the previous graph
slope m = Tan ø
Where ø is the angle the line makes with the x-axis
So the slope of the red line is
“the tangent of the mean of the angles of the slopes of the two blue lines.”
Write that using Tan-1ø!

11. Another Algebraic Viewpoint
We know speed = dist/time
So speed x time = distance
Therefore
30(t-1) = 20(t+1) which is the distance d
solving this equation gives
t=5hrs and hence d = 120km
and the correct speed of 24km/hr.

12. A Calculus Solution
The rate of change dy/dt = 30 km/hr and if this is integrated with the boundary conditions that when t=0, y=0 and when t = t-1, y = d
We get the equation
30(t-1)=d
Likewise 20(t+1)=d and we have
the same solution as the last slide.

13. Clever Algebra Solution
I know 30(t -1) = d ……(1)
and 20(t+1) = d ……(2)
If I triple eqn (2) and double eqn (1)
Then subtract I quickly get
d = 120km

14. Even more algebra! 3x eqn (2) is 60t + 60 = 3d And
When we add these two equations
We get 120t = 5d
Or 120/5=d/t=v = 24km/hr

15. Another view! If we want to arrive on time d/30 + 1 = d/20 - 1
Show this gives us the solution
d = 120km

16. Math Sense
I travel at 20km/hr for longer than I travel at 30km/hr and yet I go the same distance.
So 3hr x 20km/hr and 2hr x 30km/hr all divided by the total time (3hrs + 2 hrs) I travel;
must be the average speed
v=24km/hr.

17. More Bizzare
If I travelled there at 20km/hr and back at 30km/hr I would get home on time.
30(t - 1) + 20(t + 1) = 2d
ie 50t = 2d
25t = d
or 5(5t - 1) = d
Which means d is a multiple of 5
and the product of 5 and
one less than a multiple of 5.

18. Which means
x4 = 20
x9 = 45
x14 = 70
x19 = 95
x 24 = 120
x 29 = 145
and so on…
are the ONLY possible solutions.
120 is the first CM of 20 and 30.

19. Numbersense?
The answer is between 20 and 30 and it is closer to 20 than 30 because I spend longer travelling at 20.
It is closer in the ratio 2:3 or 4:6
The number I am looking for is 24.

20. Never only ONE way
There are many different ways to solve problems. We should seek other ways and share the results.
We must connect the table to the graph to the algebraic equation.
When we understand the properties we discover the beauty of mathematics.