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4.3.04

The Miller and Donkey problem - SOLUTION 

Are you able to discover an equation-solution for the miller and the donkey problem?

A miller intends to carry 100 bags of wheat, contends each one 100 kg, from his house up to a mill that is distant 100 km from there. For such uses a donkey that he knows not to support a load more than 100 kg. However the problem is that the donkey when loaded it needs to ingest 1 kg of wheat for each km that it covers.

The question is: Which is the biggest amount of wheat that the miller will make to arrive to the mill?

Note 1: The bags have null weight.
Note 2: The donkey without load does not consume wheat none "



SOLUTION

1. General Solution

From my point of view we can model this problem in mathematical terms as a decay phenomenon. Our task starts with identifying the variables that are changing in the phenomenon or system and a set of reasonable assumptions about them. This mathematical model involves two related variables were

x = the space walked by the donkey at every instant.

y(x) = the remaining amount of wheat at every instant.

It also involves some constants, namely:

s = distance between the miller's home and the mill.

Yo = the initial amount of wheat when x = 0

k = rate constant (fractional loss of mass per unit length)

With respect to the set of reasonable assumptions let us assume that x and y are continuous variables and linearly related. It is intuitively clear that the amount of wheat (y) decays when the donkey advances on its path.

Thus it is easy to see the rate of change of wheat y(x) with distance (x) is proportional to the ratio between them.

(1) dy/dx = - k [y(x)/s]

where k is one constant of proportionality.

By separatin the variables and integrating both sides, we have

(2) ln y(x) = - (k/s) x + c

where c is one arbitrary constant of integration.

Since

(3) y(x) = Yo when x = 0

these are the boundary conditions. Replacing the values above into (2) we get

(4) c = ln Yo

Substituting this value of c into (2) we have

(5) ln y(x) = - (k/s) x + ln Yo

Solving the equation in order to y we arrive at

(6) y(x) = Yo e^ - (k/s) x

which is the general solution of the differential equation (1).


This problem is an example of exponential decay. The positive constant k is called the rate constant , for its value is clearly a measure of the rate wich the decrease proceeds.

K can be thought of as the fractional loss of wheat per unit lenght.


In order to continue and find y as function of x, we must have further information about k. For such a purpose let us imagine the donkey path is divided in 100 equal parts. This means each part measures 1 km lenght. The miller must unload 100 bags at the end of 1st. part and to give 100 kg of wheat to the donkey since it eats 1kg per km. Thus at this point of the path 9900 kg are remaining.


2 . Particular Solution

The desired particular solution can be found from the general solution (6) by determining the value of the constant k as we did above relatively to c.

Remembering the equation (5)

ln y(x) = - (k/s) x + ln y(x)

and solving it in order to k we get

(7) k = - [ s ln [ y(x) / Yo ) / x]


Now using the fact that

y = 9 900 kg when x = 1 km

and substituing these values into (7) we get


(8) k = 1,00503350535*


* k should be equal to 1. This means the boundary condition y(1) = 9900 is not completetly exact.


Replacing this value into the general equation (6) we arrive at


3. General Solution of the Problem


(9) y(x) = Yo e^- 1,005036 x/s


* If the last boundary condition was exact the equation (9) should be

(9.1) y(x) = Yo e^- x/s

4. Particular Solution of the Problem

Since at the moment the donkey finishes its task x = s

Thus x/s = 1

so that (9) becomes


(10) y = Yo e ^ - 1,005034

wich it is the particular solution of the problem that satisfies the following boundary conditions:


y(0) = 10 000 kg

y(1) = 9 900 kg *

* We saw above that this condition is not completelly exact.


Now placing the numerical values into (10) we finally arrive at desired answer:


y(s) = 10 000 e^ -1,00503350535


y(s) = 3 660,323 kg



Analytical result = 3 660,323 kg* (approximate result)

Computer result = 3647,427 kg (correct result)

Difference = 12,896 kg

Precision = 99,65%


* If k =1 then y(s) = Yo/e , that is, 3678,794 kg.

5. Conclusion

Before finishing I must add that in most cases a mathematical model is only an approximation of the physical condition being studied. In our problem, for example, for a specified increment of x (distance) the mass of wheat decreases by discrete amounts.

This mathematical model assumes that the decay process is a continuous function of the distance walked. In spite of these restrictions we have obtained a good approximation (99,65%) relatively to the theoretic result.

Manuel da Cruz de Sousa

Email: manuelcruzsousa@sapo.pt

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