0. Code

>	restart; with(plots):

Warning, the name changecoords has been redefined

>

WorkTogether  :=  proc( a, b )      local x, t, A, B, pa, pb, pta, ptb;      t := solve( x/a + x/b = 1, x):      A := t/a;        B := t/b:      pa := polygonplot( [ [0,0],[A,0],[A,1],[0,1] ], color = gold, axes = none):      pb := polygonplot( [ [A,0],[1,0],[1,1], [A,1] ], color = red):      pta := textplot( [ 0, 1.2, a]);      ptb := textplot( [ 1, 1.2, b]);      display( pa, pb, pta, ptb)  end proc:

>

WorkTogetherEarly  :=  proc( a, b, c )      local pc, x, t, A, B, C, pa, pb, pta, ptb;      t := solve( x/a + x/b = 1, x):      C := c/a;      A := (1-C)*t/a;     B := (1-C)*t/b:        pc := polygonplot( [ [0,0],[C,0],[C,1],[0,1] ],                                 color = gold, axes = none):      pa := polygonplot( [ [C,0],[C+A,0],[C+A,1],[C,1] ],                                 color = gold, axes = none):      pb := polygonplot( [ [A+C,0],[1,0],[1,1], [A+C,1] ], color = red):      pta := textplot( [ A, 1.2, a]);      ptb := textplot( [ 1, 1.2, b]);      display( pc, pa, pb, pta, ptb)  end proc:

1. The General Idea

It will help us have a better feel for what is going on, if we can invision the work done by each party.

Problem   1.1 :  Rick takes 3 hours to complete a job working alone. Stewart takes four hours. How long will it take if they work together?

The person who can complete the job in less time is actually the faster worker - the person who will get more done in the same amount of time.

>	WorkTogether( 3, 4);

If a person takes 10 hours to do the job, compared to someone who can do it one hour, that first person gets fairly little done.

>	WorkTogether( 10, 1);

It's even worse if the first person takes 100 hours to do the job.

>	WorkTogether( 100, 1);

But it's the other way around if the first person takes an hour, while the 2nd takes 100 hours.

>	WorkTogether( 1, 100);

>

2. Solving Simple Work Problems

  Problem  2.1:  Rick takes 3 hours to complete a job working alone. Stewart takes four hours. How long will it take if they work together?

Rick works at a rate of 1/3 job/hour, while Stewart works at a rate of 1/4 of a job/hour. If they both work for t hours, then this is how much work they will accomplish.

>	t/3 + t/4;

If we take this expression, which represents the amount of work they do, and set it equal to one whole job, we get :

>	t/3 + t/4 =1;

>	t = solve( %, t);

Problem  2.2:   A storage tank has two input pipes, A and B. Pipe A can fill the the tank in 20 hours by itself. Pipe B can fill the tank in 35 hours by itself. If both pipes are turned on at the same time, how long will it take to fill the tank?

Pipe A has a rate of 1/20 of a tank per hour, or t/20 of a tank in t hours.
Pipe A has a rate of 1/35 of a tank per hour, or t/35 of a tank in t hours.

>	t/20; t/35;

If add how much is accomplished by pipe A and how much by B, and add them together, we find out how much of the tank is full after t hours. In particular we want the sum of the output to equal 1 full tank.

>	t/20 + t/35 = 1;

>	t = solve( %, t); evalf(%);

3. Graphic View of More Complicated Work Problems

 
Problem 3.1 :  Alanis can complete a job in 7 hours if working alone. Britany can complete the job in 10 hours if working alone. If Alanis works for 2 hours by herself, and then is joined by Britany, how much faster can the job be done?

>	WorkTogetherEarly( 7, 10, 2);

What if Alanis worked for 4 hours before Britany came on board?

>	WorkTogetherEarly( 7, 10, 4);

What if Alanis works for a full 6 hours, before Britany shows up?

>	WorkTogetherEarly( 7, 10, 6);

4. The General Idea of More Complicated Work Problems

Problem 4.1 :  Alanis can complete a job in 7 hours if working alone. Britany can complete the job in 10 hours if working alone. If Alanis works for 2 hours by herself, and then is joined by Britany, how much quicker can the job be done?

Alanis' rate is 1/7 job/hour. If she works for 2 hours at that rate she will complete this much of the job :

>	2 * (1/7);

This is how much of the job remains to be done, when Britany starts.

>

1 - %;

When graphing these lines, the best and fastest method is to find the intercepts and then 'connect the dots'

>	t/7 + t/10 = %;

Here are other examples.

>	answer := solve( %, t);

Check the answer by putting this number back in. Here is how much of the job Alanis gets done :

>	(2 + answer) * (1/7);

Here is how much of the job Britany does.

>	answer * (1/10);

Together, these two portions of the job should add up to the entire job. If these are correct, we should find that they add to one.

>

% + %%;