Mini-Project
for
Calculus and Mathematica
Workshop

Clear[x,y]
  Plot3D[Sin[x y]+1,
        {x,-Pi, Pi},{y, - Pi, Pi},
         PlotPoints->65,
         Lighting->True,
         Mesh->False,
         Ticks->None,
         Boxed->False,
         Axes->None];

Name:     Shirley Barrette

Title:       Just Another Job

Description:   This is a calculus problem concerning "on the  job" teaming.   As workers are teamed together, they tend to interfere with one another.

JUST ANOTHER JOB

Workers are being hired in your factory to form a team which will perform a certain job.  Each worker, working alone, can complete the task in 30 hours.  As workers are teamed together, they tend to interfere with one another.  Alone, each member works at the rate of completing 1/30 of the job per hour but with other workers this individual rate lessens.  Motion study experts have discovered the following interference pattern in your factory.  If two workers are grouped, each one's rate of doing work lessens in such a way that at the slower rate of the group, each individual would require 6% more time to do the job if they were working alone.  That is, each would require (30)(1.06) more time to complete the job alone.  If three workers are in a team, the individual rate of each worker lessens further by a compounded amount.  In comparison to the two-member team, each person in the three-member team would require an additional 6%  more time to do the job working alone than they would have as members of a two-member team.  In other words, each worker in the three-member team would require (30)(1.06)¤ hours to complete the job alone.  This pattern continues in exactly the same way as each additional member is added to the team.

1.  How many workers should be placed in a team so that the job  is done in the least possible time by the team?
2.  What is this least time?
3.  Justify your answer by proving that it does result in a minimum working time.

ONE CALCULUS SOLUTION

If we let x represent the number of workers on the team, then each worker works at such a rate that each alone would require (30)(1.06)Ù—∕ hours.  Then the rate of each of the x workers is 1/(30)(1.06)Ù—∕ of the job done per hour.  If we let T[x] be the time for the whole team of x workers to do one job, the fraction of the job done by one of the workers in T[x] hours is T[x]/(30)(1.06)Ù—∕.

Now we have the equation:

(x)T[x]/(30)(1.06)Ù—∕ = 1

Okay, solve this equation for T[x].

T[x] = ((30)(1.06)Ù—∕)/x
 
 LET'S GRAPH THIS BABY!

Clear[x]
T[x_] = ((30)(.06)^x-1)/x
Plot[T[x],{x,1,20},
PlotRange->{-5,1},
PlotStyle->{RGBColor[0,1,0]}];

It appears that this function approaches 0 somewhere between 15 and 20.

Use the logarithmic differentiation to find T'[x].

Log[T[x]]=Log[30] + (x-1)Log[1.06] - Log[x]
T[x_]=30 (1.06)^(x-1) /x
Log[T[x_]] = Log[30] + (x-1) Log[1.06] - Log[x]
T'[x]/T[x] = Log(1.06) - 1/x
If we set T'[x] = 0, we get the result, x = 1/Log(1.06)

N[1/.0582689081239757753]

AH HA, 	so x = 17.1618.  Let's round to 17.2.

To verify that this value for x minimizes T[x], find T''[x].

T'[x] = T[x]{Log(1.06)-x—∕}
T''[x] = T'[x]{Log(1.06)-x—∕} + T[x]/x¤
T''[x] = {Log(1.06) - x—∕)¤ + 1/x¤

Since T''[1/Log(1.06)] > 0, T[1/Log(1.06)] is the minimum.  
We know that Log 1.06 = 0.0582689081239757753

N[1/.0582689081239757753]

WHAT CAN I SAY?????  
IF 17 WORKERS FORM A TEAM, 
THE WORKING TIME FOR THE TEAM IS A MINIMUM!!!!
UGH, WHAT A ROCK!!  
I really didn't care much for that nasty brain-boggling stuff.  

Let's see if we can use Mathematica instead.

ANOTHER WAY WITH Mathematica

s=Solve[(x) T[x]/((30) (1.06)^(x-1))==1,T[x]]

That's not the format we want for T[x].  Enter the following.

T[x_]=s[[1,1,2]]

That's better.

Let's look at the rate of change for T'[x].

T'[x]

Now we need to solve T'[x] = 0

Solve[T'[x]==0,x]

WHOA! GROSS!  Mathematica can not do this for you.  What to do?  What to do? 
Let's Plot T'[X].

Clear[x]
Plot[T'[x],{x,1,20},
PlotRange->{-5,1},
PlotStyle->{RGBColor[0,1,0]}];

Look at that great graph. You can see that this function approaches the x-axis somewhere between 15 and 20.  Let's use FindRoot.

FindRoot[T'[x],{x,15}]

HOORAY!! We need 17 workers so that the job can be done in the least possible time by the team.

GUESS AND CHECK

We could also use GUESS AND CHECK to find the number of workers needed.

Using T'[x], GUESS VALUES FOR x AND CHECK THEM, until you find the minimum number of workers needed to do the job.  Remember, we want T'[x] to be as close to 0 as possible.

T'[x]

Don't you think that you should say,
"Thank you Mathematica for an easier, faster way 
to find the solution to my problem and MAKE MY DAY."
Signing off for now, MATHELETES!
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